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Circuitos electrónicos_Sánchez_EPN
Viviana López
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Vista previa del material en texto
Seguramente no hay nada más contemporáneo que la tec-
nología de la semiconducción y los diferentes dispositivos que
con ella se han logrado. Estudiarlos y esforzarnos por entender
el funcionamiento de los complejos sistemas que han logrado
constituye, pues, una de las actividades intelectuales más
fructíferas y urgentes en nuestros días. El presente libro está
dedicado precisamente a ello.
Los dispositivos electrónicos actúan unos con otros para
formar circuitos y estos, a la vez, forman redes circuitales más
complejas que, obedeciendo a leyes y principios eléctricos,
permiten brindar soluciones a requerimientos y problemas de
ingeniería. La electrónica actual se enfoca en la búsqueda de
nuevos materiales para el diseño y la manufactura de nuevos
dispositivos y circuitos electrónicos utilizados en la elaboración
de aparatos, objetos e instrumentos que mejoran la calidad
de vida de las personas. En tal contexto, esta obra recoge la
fundamentación teórica, ofrece sendos ejemplos y ejercicios
propuestos, así como también presenta algunas de las apli-
caciones prácticas de los circuitos electrónicos. La exposición
de estos materiales está organizada progresivamente en tres
partes. En suma, Circuitos electrónicos: ejercicios y aplicaciones
busca ser un útil insumo para los estudiantes de ingeniería.
Tarquino Sánchez Almeida. Ingeniero en Electrónica y Telecomunicaciones y MBA
en Gerencia Empresarial por la Escuela Politécnica Nacional (EPN) y en la Maastricht
School of Management (Países Bajos) obtuvo un Post Graduate Diploma. Ha sido
decano, subdecano, coordinador de posgrados y docente investigador de la Facultad
de Ingeniería Eléctrica y Electrónica de la EPN. Además, se ha desempeñado como
docente invitado en las universidades de Cuenca, De las Américas e Internacional del
Ecuador. También es autor de varias publicaciones científicas en revistas indexadas y
de libros de apoyo a la docencia. Actualmente es Vicerrector de Docencia de la Escuela
Politécnica Nacional (2013-2018).
para mayor información, visite:
www.epn.edu.ec
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Circuitos
ejercicios y aplicaciones
electrónicos
Tarquino Sánchez Almeida
ISBN 978-9978-383-27-8
EstebanCrespo
Typewritten Text
EstebanCrespo
Typewritten Text
𝐶𝐵 𝐶𝑐
𝑉𝑐𝑐 (𝐶𝐵) 𝑅𝐿 (𝐶𝑐)
𝐶𝐸
𝑅𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑖𝑎 𝑑𝑖𝑛á𝑚𝑖𝑐𝑎 𝑑𝑒 𝑙𝑎 𝐽𝐵𝐸 𝑟𝑒 =
𝑉𝑇
𝐼𝐸
𝑉𝑜𝑙𝑡𝑎𝑗𝑒 𝑡é𝑟𝑚𝑖𝑐𝑜 𝑉𝑇 =
𝐾 ∗ 𝑇
𝑞
𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑑𝑒 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝐾 = 1.38 ∗ 10−23
𝐽
𝐾
𝐶𝑎𝑟𝑔𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑡𝑟ó𝑛 𝑞 = 1.6 ∗ 10−19 𝐶
𝑟𝑒 =
26 𝑚𝑉
𝐼𝐸
𝑟𝑑 = 𝑑𝑒𝑐𝑒𝑛𝑎𝑠 𝑑𝑒 𝛺 (𝑑𝑒𝑠𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒)
𝑟𝐶 , 𝑟𝐶
′ = 𝑑𝑒𝑐𝑒𝑛𝑎𝑠 𝑜 𝑐𝑒𝑛𝑡𝑎𝑛𝑎𝑠 𝑑𝑒 𝑀𝛺 (𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑜 𝑎𝑏𝑖𝑒𝑟𝑡𝑜)
𝑟𝐶 , 𝑟𝐶
′ → ∞
𝑟𝑏 ≈ 0 𝛺
𝑅𝐿
′ = 𝑅𝐶‖𝑅𝐿 [1.1]
𝑅𝐿
(𝐴𝑖)
𝑖𝑐 𝑖𝐵
𝐴𝐼 =
𝑖𝐶
𝑖𝐵
= ℎ𝑓𝑒 = 𝛽 [1.2]
(𝐴𝑣)
𝐴𝑣 =
𝑉𝑂
𝑉𝑖𝑛
≈ −
𝛽 ∗ 𝑅𝐿
′
(𝛽 + 1) ∗ 𝑟𝑒
𝑆𝑖 𝛽 ≫ 1 → 𝛽 ≈ 𝛽 + 1
𝐴𝑣 ≈ −
𝑅𝐿
′
𝑟𝑒
[1.3]
𝑍𝑖𝑛𝑇 ≅ ℎ𝑖𝑒 ≈ (𝛽 + 1) ∗ 𝑟𝑒 [1.4]
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇‖ 𝑅𝐵𝐵 [1.5]
𝑅𝐵𝐵 = 𝑅𝐵1 ‖ 𝑅𝐵2
𝑍𝑜𝑇 >
1
ℎ𝑜𝑒
[1.6]
𝑍𝑂 = 𝑍𝑜𝑇‖𝑅𝐿
′
𝐶𝑜𝑚𝑜 𝑍𝑜𝑇 ≫ 𝑅𝐿
′
𝑍𝑂 = 𝑅𝐿
′ [1.7]
CE
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒 + 𝑋𝐶𝐸‖𝑅𝐸
𝑋𝐶𝐸 ≪ 𝑅𝐸
𝑋𝐶𝐸‖𝑅𝐸 ≈ 𝑋𝐶𝐸
𝑋𝐶𝐸 ≪ 𝑟𝑒
𝐶𝐸 ≥
10
𝑤 ∗ 𝑟𝑒
[1.8]
𝐂𝐄
𝐴𝐼 =
𝑖𝐶
𝑖𝐵
= ℎ𝑓𝑒 = 𝛽 [1.9]
𝐴𝑣 =
𝑉𝑂
𝑉𝑖𝑛
≈ −
𝛽 ∗ 𝑅𝐿
′
(𝛽 + 1) ∗ (𝑟𝑒 + 𝑅𝐸)
𝑆𝑖 𝛽 ≫ 1 → 𝛽 ≈ 𝛽 + 1
𝐴𝑣 ≈ −
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
[1.10]
𝑍𝑖𝑛𝑇 ≅ ℎ𝑖𝑒 ≈ (𝛽 + 1) ∗ (𝑟𝑒 + 𝑅𝐸) [1.11]
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇‖ 𝑅𝐵𝐵 [1.12]
𝑅𝐵𝐵 = 𝑅𝐵1 ‖ 𝑅𝐵2
𝑍𝑜𝑇 >
1
ℎ𝑜𝑒
[1.13]
𝑍𝑂 = 𝑍𝑜𝑇‖𝑅𝐿
′
𝐶𝑜𝑚𝑜 𝑍𝑜𝑇 ≫ 𝑅𝐿
′
𝑍𝑂 = 𝑅𝐿
′ [1.14]
𝑟𝑒
𝑅𝐸
𝑅𝐸 = 𝑅𝐸
′ + 𝑅𝐸
′′
𝑅𝐸 = 𝑅𝐸
′
𝐴𝑣 ≈ −
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ [1.15]
𝑍𝑖𝑛𝑇 = (𝛽 + 1) ∗ (𝑟𝑒 + 𝑅𝐸
′ ) [1.16]
CE
𝟏. 𝑋𝐶𝐸 ≪ 𝑅𝐸
𝑋𝐶𝐸||𝑅𝐸 ≈ 𝑋𝐶𝐸
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ + 𝑋𝐶𝐸
𝟐. 𝑋𝐶𝐸 ≪ 𝑟𝑒 + 𝑅𝐸
′
𝑆𝑒 𝑡𝑜𝑚𝑎 𝑙𝑎 𝑐𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 2 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝑞𝑢𝑒: 𝑟𝑒 + 𝑅𝐸
′ < 𝑅𝐸
𝐶𝐸 ≥
10
𝜔(𝑟𝑒 + 𝑅𝐸
′ )
[1.17]
𝑟𝑒
𝑉𝐶𝐸 = �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 , [1.18]
𝑉𝐶𝐸𝑚𝑖𝑛 𝑉𝐶𝐸
𝑉𝐶𝐸𝑠𝑎𝑡 = 0.2 𝑉
𝑉𝐶𝐸𝑚𝑖𝑛 ≫ 𝑉𝐶𝐸𝑠𝑎𝑡
𝑉𝐶𝐸𝑚𝑖𝑛 ≈ 2 𝑉 [1.19]
𝐼𝐶𝑄 ≥ 𝐼𝐶
𝐼𝐶𝑅𝐿
′ ≥ �̂�𝑂
𝑉𝑅𝐶
𝑅𝐶
≥
�̂�𝑂
𝑅𝐿
′
𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅𝐿
′ �̂�𝑜 [1.20]
𝐼𝐸𝑅𝐸
′ ≥ �̂�𝑖𝑛
𝑉𝐸
𝑅𝐸
≥
𝑉𝑖𝑛
𝑅𝐸
′
𝑉𝐸 ≥
𝑅𝐸
𝑅𝐸
′ �̂�𝑖𝑛 [1.21]
𝑉𝑐𝑐
𝑉𝐶𝐶 ≥ 𝑉𝐸 + 𝑉𝐶𝐸 + 𝑉𝑅𝐶 [1.22]
𝑉𝐶𝐶 ≥ 𝑉𝐸 + �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 + 𝑉𝑅𝐶
[1.23]
𝑉𝐶𝐶 ≈ 1.1𝑉𝐶𝐶 [1.24]
𝑉𝑖𝑛 = 0.5 sen 𝜔𝑡 V 1 kΩ
𝛽𝑚𝑖𝑛 = 50, 𝛽𝑡í𝑝𝑖𝑐𝑜 = 80, 𝛽𝑚á𝑥 = 100. 1 kHz
20 kHz
𝑅𝐶 ≫ 𝑅𝐿 ⟹ 𝑅𝐿
′ = 𝑅𝐿
𝑅𝐶 = 𝑅𝐿 ⟹ 𝑅𝐿
′ =
𝑅𝐿
2
=
𝑅𝐶
2
𝑅𝐶 ≪ 𝑅𝐿 ⟹ 𝑅𝐿
′ = 𝑅𝐶
|𝐴𝑣| = 10
𝛽𝑚𝑖𝑛 = 50
𝑉𝑖𝑛 = 0.5 𝑆𝑒𝑛(𝜔𝑡) 𝑉
𝑓𝑚𝑖𝑛 = 1 𝑘𝐻𝑧
𝑅𝐿 = 1 𝑘𝛺
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑅𝐶 > 𝑅𝐿
𝑹𝑪 = 𝟐 𝒌𝜴
�̂�𝑜 = |𝐴𝑣| × 𝑉𝑖𝑛 = 10 × 0.5 𝑉 = 5 𝑉
𝑅𝐿 = 1 𝑘𝛺
𝑅𝐿
′ = 𝑅𝐶 ∥ 𝑅𝐿 = 2 𝑘Ω ∥ 1 𝑘Ω = 666.67 Ω
𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅𝐿
′ 𝑉�̂� =
2 𝑘𝛺
666.67 Ω
(5 𝑉)(1.1) = 16.5 𝑉
1.1 = 𝐹𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
𝐼𝐶 =
𝑉𝑅𝐶
𝑅𝐶
=
16.5 𝑉
2 𝑘Ω
= 8.25 𝑚𝐴 = 𝐼𝐸
1 𝑚𝐴 < 𝐼𝐶 < 10 𝑚𝐴
𝐿𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝐼𝐶 𝑐𝑢𝑚𝑝𝑙𝑒 𝑐𝑜𝑛 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑒𝑙 𝑇𝐵𝐽.
𝐼𝐵 =
𝐼𝐶
𝛽𝑚𝑖𝑛
=
8.25 𝑚𝐴
50
= 165 µ𝐴
𝑉𝑇 =
𝐾𝑇𝐾
𝑞
=
(1.3806503 × 10−23
𝐽
𝐾) (300 𝐾)
1.6021765 × 10−19 𝐽
= 25.852 𝑚𝑉
𝑑𝑜𝑛𝑑𝑒:
𝑉𝑇 = 𝑣𝑜𝑙𝑡𝑎𝑗𝑒 𝑡é𝑟𝑚𝑖𝑐𝑜
𝑇𝐾 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑎𝑚𝑏𝑖𝑒𝑛𝑡𝑒
𝐾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑑𝑒 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛
𝑞 = 𝑐𝑎𝑟𝑔𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑐𝑡𝑟ó𝑛
𝑃𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜 𝑠𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑉𝑇 = 26 𝑚𝑉
𝑟𝑒 =
𝑉𝑇
𝐼𝐶
=
26 𝑚𝑉
8.25 𝑚𝐴
= 3.15 Ω
𝑅𝐸
′ =
𝑅𝐿
′
|𝐴𝑉|
−𝑟𝑒 =
666.67 Ω
10
− 3.15 Ω = 63.52 𝛺
↑ 68 Ω
↓ 62 Ω
𝑹𝑬
′ = 𝟔𝟐 𝜴 𝒑𝒂𝒓𝒂 𝒄𝒖𝒎𝒑𝒍𝒊𝒓 𝒍𝒂 𝒈𝒂𝒏𝒂𝒏𝒄𝒊𝒂
𝑉𝐶𝐸 ≥ �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 ≥ 5 𝑉 + 0.5 𝑉 + 2 𝑉 ≥ 7.5 𝑉
𝑉𝐶𝐸 = 8 𝑉
𝑉𝑅𝐵1 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 − 𝑉𝐸𝐵 = 16.5 𝑉 + 8 𝑉 − 0,7 𝑉 = 23.8 𝑉
𝐼1 = 11 × 𝐼𝐵 = 11(165 µ𝐴) = 1.82 𝑚𝐴
𝑅𝐵1 =
𝑉𝑅𝐵1
𝐼1
=
23.8 𝑉
1.82 𝑚𝐴
= 13.07 𝑘Ω
↑ 15 𝑘Ω
↓ 13 𝑘Ω
𝑹𝑩𝟏 = 𝟏𝟑 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝐼1 =
𝑉𝑅𝐵1
𝑅𝐵1
=
23.8 𝑉
13 𝑘Ω
= 1.83 𝑚𝐴
𝐼2 = 𝐼1 − 𝐼𝐵 = 1.83 𝑚𝐴 − 165 µ𝐴 = 1.67 𝑚𝐴
𝑉𝐸 ≥ 1 𝑉 + �̂�𝑖𝑛 ≥ 1 𝑉 + 0.5 𝑉 ≥ 1.5 𝑉
𝑉𝐸 = 2 𝑉
𝑅𝐸 =
𝑉𝐸
𝐼𝐸
=
2 𝑉
8.25 𝑚𝐴
= 242.42 𝛺
𝑅′𝐸
′ = 𝑅𝐸 − 𝑅𝐸
′ = 242.42 𝑘𝛺 − 62 𝛺 = 180.42 𝛺
↑ 200 Ω
↓ 180 Ω
𝑹𝑬
′′ = 𝟏𝟖𝟎 𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐵 = 𝑉𝐸 + 𝑉𝐵𝐸 = 2 𝑉 + 0.7 𝑉 ≥ 2.7 𝑉
𝑅𝐵2 =
𝑉𝐵
𝐼2
=
2.7 𝑉
1.67 𝑚𝐴
= 1.61 𝑘𝛺
↑ 1.8 𝑘Ω
↓ 1.6 𝑘Ω
𝑹𝑩𝟐 = 𝟏. 𝟔 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐶𝐶 ≥ 𝑉𝑅𝐶 + 𝑉𝐶𝐸 + 𝑉𝐸 ≥ 16.5 𝑉 + 8 𝑉 + 2 𝑉 ≥ 26.5 𝑉
𝑽𝑪𝑪 = 𝟐𝟕 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝑍𝑖𝑛−𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐸
′ ) = (50 + 1)(3.15 𝛺 + 62 𝛺) = 3.32 𝑘Ω
𝑍𝑖𝑛 = 𝑅𝐵1 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛−𝑇 = 1.6 𝑘Ω ∥ 13 𝑘Ω ∥ 3.32 𝑘𝛺 = 996.88 Ω
|𝐴𝑣| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ =
666.67 Ω
3.15 𝛺 + 62 𝛺
= 10.23
𝐶𝐶 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 666.67 Ω
≥ 2.39 µ𝐹
𝑪𝑪 = 𝟑. 𝟑 µ𝑭
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸
′ )
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × (3.15 Ω + 62 Ω)
= 24.43µ𝐹
𝑪𝑬 = 𝟑𝟑 µ𝑭
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 996.88 Ω
= 1.59 µ𝐹
𝑪𝑩 = 𝟐. 𝟐 µ𝑭
|𝐴𝑣| = 40
𝛽𝑚𝑖𝑛 = 50
𝑉𝑖𝑛 = 0.1 𝑆𝑒𝑛(𝜔𝑡) 𝑉
𝑓𝑚𝑖𝑛 = 1 𝑘𝐻𝑧
𝑅𝐿 = 3.3 𝑘𝛺
𝑍𝑖𝑛 ≥ 1 𝑘𝛺
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑚á𝑥𝑖𝑚𝑎 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜:
|𝐴𝑣| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′
(𝑟𝑒 + 𝑅𝐸
′ )𝑚á𝑥 =
𝑅𝐿
′
𝑚á𝑥
𝐴𝑣
=
𝑅𝐿
𝐴𝑣
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑚í𝑛𝑖𝑚𝑎 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜:
𝑍𝑖𝑛 = 𝑅𝐵𝐵 ∥ 𝑍𝑖𝑛𝑇
𝑍𝑖𝑛𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐸
′ )
𝑆𝑖 𝑅𝐵𝐵 ≫ 𝑍𝑖𝑛𝑇
(𝑟𝑒 + 𝑅𝐸
′ )𝑚𝑖𝑛 =
𝑍𝑖𝑛𝑇𝑚𝑖𝑛
(𝛽 + 1)
=
𝑍𝑖𝑛
(𝛽 + 1)
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜:
𝑍𝑖𝑛
𝛽 + 1
< 𝑟𝑒 + 𝑅𝐸
′ <
𝑅𝐿
|𝐴𝑣|
¿ 𝐸𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑖𝑠𝑒ñ𝑎𝑟 𝑐𝑜𝑛 𝑒𝑠𝑡𝑎𝑠 𝑐𝑜𝑛𝑑𝑖𝑐𝑖𝑜𝑛𝑒𝑠?
1 𝑘𝛺
50 + 1
< 𝑟𝑒 + 𝑅𝐸
′ <
3.3 𝑘𝛺
40
19.61 𝛺 < 𝑟𝑒 + 𝑅𝐸
′ < 82.5 𝛺
𝑆í 𝑒𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑖𝑠𝑒ñ𝑎𝑟
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑟𝑒 + 𝑅𝐸
′ = 40 𝛺
|𝐴𝑣| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′
𝑅𝐿
′ = |𝐴𝑣|(𝑟𝑒 + 𝑅𝐸
′ ) = (40)(40 𝛺) = 1.6 𝑘𝛺
𝑅𝐶 =
𝑅𝐿 × 𝑅𝐿
′
𝑅𝐿 − 𝑅𝐿
′ =
3.3 𝑘Ω × 1.6 𝑘𝛺
3.3 𝑘𝛺 − 1.6 𝑘Ω
= 3.11 𝑘Ω
↑ 3.3 𝑘Ω
↓ 3.0 𝑘Ω
𝑹𝑪 = 𝟑 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
�̂�𝑜 = |𝐴𝑣| × �̂�𝑖𝑛 = 40 × 0.1 𝑉 = 4 𝑉
𝑅𝐿 = 3.3 𝑘𝛺
𝑅𝐿
′ = 𝑅𝐶 ∥ 𝑅𝐿 = 3 𝑘𝛺 ∥ 3.3 𝑘𝛺 = 1.57 𝑘𝛺
𝑉𝑅𝐶 =
𝑅𝐶
𝑅𝐿
′ × �̂�𝑜 =
3 𝑘𝛺
1.57 𝑘𝛺
(4 𝑉)(1.1) = 8.41 𝑉
1.1 = 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
𝐼𝐶 =
𝑉𝑅𝐶
𝑅𝐶
=
8.41 𝑉
3 𝑘Ω
= 2.80 𝑚𝐴 = 𝐼𝐸
1 𝑚𝐴 < 𝐼𝐶 < 10 𝑚𝐴
𝐿𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝐼𝐶 𝑐𝑢𝑚𝑝𝑙𝑒 𝑐𝑜𝑛 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑒𝑙 𝑇𝐵𝐽.
𝐼𝐵 =
𝐼𝐶
𝛽𝑚𝑖𝑛
=
2.80 𝑚𝐴
50
= 56 µ𝐴
𝑉𝑇 =
𝐾𝑇𝐾
𝑞
=
(1.3806503 × 10−23
𝐽
𝐾) (300 𝐾)
1.6021765 × 10−19 𝐽
= 25.852 𝑚𝑉
donde:
𝑉𝑇 = 𝑣𝑜𝑙𝑡𝑎𝑗𝑒 𝑡é𝑟𝑚𝑖𝑐𝑜
𝑇𝐾 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑎𝑚𝑏𝑖𝑒𝑛𝑡𝑒
𝐾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑑𝑒 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛
𝑞 = 𝑐𝑎𝑟𝑔𝑎 𝑑𝑒𝑙 𝑒𝑙𝑒𝑐𝑡𝑟ó𝑛
𝑃𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜 𝑠𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑉𝑇 = 26 𝑚𝑉
𝑟𝑒 =
𝑉𝑇
𝐼𝐶
=
26 𝑚𝑉
2.80 𝑚𝐴
= 9.29 Ω
𝑅𝐸
′ =
𝑅𝐿
′
|𝐴𝑉|
− 𝑟𝑒 =
1.57 𝑘𝛺
40
− 9.29 Ω = 29.96 𝛺
↑ 30Ω
↓ 27Ω
𝑹𝑬
′ = 𝟑𝟎 𝜴 𝒑𝒂𝒓𝒂 𝒄𝒖𝒎𝒑𝒍𝒊𝒓 𝒍𝒂 𝒈𝒂𝒏𝒂𝒏𝒄𝒊𝒂
𝑉𝐶𝐸 ≥ �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 ≥ 4 𝑉 + 0.1 𝑉 + 2 𝑉 ≥ 6.1 𝑉
𝑉𝐶𝐸 = 6.5 𝑉
𝑉𝑅𝐵1 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 − 𝑉𝐸𝐵 = 8.41 𝑉 + 6.5 𝑉 − 0.7 𝑉 = 14.21 𝑉
𝐼1 = 11 × 𝐼𝐵 = 11(56 µ𝐴) = 0.62 𝑚𝐴
𝑅𝐵1 =
𝑉𝑅𝐵1
𝐼1
=
14.21 𝑉
0.62 𝑚𝐴
= 22.92 𝑘Ω
↑ 24 𝑘Ω
↓ 22 𝑘Ω
𝑹𝑩𝟏 = 𝟐𝟒 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝐼1 =
𝑉𝑅𝐵1
𝑅𝐵1
=
14.21 𝑉
24 𝑘𝛺
= 0.59 𝑚𝐴
𝐼2 = 𝐼1 − 𝐼𝐵 = 0.59 𝑚𝐴 − 56 µ𝐴 = 0.53 𝑚𝐴
𝑍𝑖𝑛 = 𝑅𝐵𝐵 ∥ 𝑍𝑖𝑛−𝑇 ≥ 1 𝑘𝛺
𝑍𝑖𝑛−𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐸
′ ) = (50 + 1)(9.29 𝛺 + 30 𝛺) = 2.0 𝑘𝛺
𝑅𝐵𝐵 =
𝑍𝑖𝑛 × 𝑍𝑖𝑛−𝑇
𝑍𝑖𝑛−𝑇 − 𝑍𝑖𝑛
=
(1 𝑘Ω)(2.0 𝑘Ω)
2.0 𝑘Ω − 1 𝑘Ω
= 2.0 𝑘Ω
𝑅𝐵2 >
𝑅𝐵𝐵 × 𝑅𝐵1
𝑅𝐵1 − 𝑅𝐵𝐵
>
(2.0 𝑘Ω)(24 𝑘Ω)
24 𝑘𝛺 − 2.0 𝑘Ω
> 2.18 𝑘Ω
↑ 2.2 𝑘Ω
↓ 2.0 𝑘Ω
𝑅𝐵2 = 2.2 𝑘Ω 𝑝𝑜𝑟 𝑐𝑒𝑟𝑐𝑎𝑛í𝑎
𝑆𝑒 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎: 𝑉𝐵 ≥ 𝑉𝐸 + 𝑉𝐵𝐸 ≥ 1 𝑉 + 0.7 ≥ 1.7 𝑉
𝑉𝐵 = 𝑅𝐵2 × 𝐼2 = (2.2 𝑘Ω)(0.53 𝑚𝐴) = 1.17 𝑉 ≱ 1.7 𝑉
𝑆𝑖 𝑉𝐵 𝑛𝑜 𝑐𝑢𝑚𝑝𝑙𝑒 𝑠𝑒 𝑒𝑠𝑐𝑜𝑗𝑒 𝑢𝑛 𝑅𝐵2 𝑚𝑎𝑦𝑜𝑟
𝑉𝐵 = 𝑅𝐵2 × 𝐼2 = (3.9 𝑘Ω)(0.53 𝑚𝐴) = 2.07 𝑉 ≥ 1.7 𝑉
𝑹𝑩𝟐 = 𝟑. 𝟗 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 2.07 𝑉 − 0.7 𝑉 = 1.37 𝑉
𝑆𝑒 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎: 𝑉𝐸 ≥ 𝑉𝑖𝑛 + 1 𝑉 ≥ 1.1 𝑉
𝑅𝐸 =
𝑉𝐸
𝐼𝐸
=
1.37 𝑉
2.80 𝑚𝐴
= 489.29 𝛺
𝑅𝐸
′′ = 𝑅𝐸 − 𝑅𝐸
′ = 489.29 𝛺 − 30 𝛺 = 459.29 Ω
↑ 470 Ω
↓ 430 Ω
𝑹𝑬
′′ = 𝟒𝟕𝟎 𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐸 = 𝐼𝐸 × 𝑅𝐸 = 𝐼𝐸 × (𝑅𝐸
′ + 𝑅𝐸
′′) = (2.80 𝑚𝐴)(30 𝛺 + 470 𝛺) = 1.4 𝑉
𝑉𝐶𝐶 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 + 𝑉𝐸 = 8.41 + 6.5 𝑉 + 1.4 𝑉 = 16.31 𝑉
𝑽𝑪𝑪 = 𝟏𝟖 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
|𝐴𝑉| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ =
1.57 𝑘Ω
9.29 𝛺 + 30 𝛺
= 39.96
𝑍𝑖𝑛 = 𝑅𝐵1 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛−𝑇 = 24 𝑘Ω ∥ 3.9 𝑘Ω ∥ 2.0 𝑘Ω = 1.25 𝑘𝛺
𝐶𝐶 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.57 𝑘Ω
≥ 1.01 µ𝐹
𝑪𝑪 = 𝟐. 𝟐µ𝑭
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸
′ )
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × (9.29 𝛺 + 30 Ω)
= 40.51µ𝐹
𝑪𝑬 = 𝟒𝟕 µ𝑭
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.25 𝑘Ω
= 1.27 µ𝐹
𝑪𝑩 = 𝟐. 𝟐 µ𝑭
A partir del desarrollo anterior se obtuvo:
𝑅𝐶 = 3 𝑘𝛺
�̂�𝑜 = 4 𝑉
𝑅𝐿 = 3.3 𝑘𝛺
𝑅𝐿
′ = 1.57 𝑘𝛺
𝑉𝑅𝐶 = 8.41 𝑉
𝐼𝐶 = 2.80 𝑚𝐴 = 𝐼𝐸
𝐼𝐵 = 56 µ𝐴
𝑟𝑒 = 9.29 𝛺
𝑅𝐸
′ = 30 Ω
𝑉𝐶𝐸 = 6.5 𝑉
𝑍𝑖𝑛−𝑇 = 2.0 𝑘𝛺
Aplicando el método 2:
𝐴𝑠𝑢𝑚𝑖𝑟: 𝑉𝐸 ≥ 𝑉𝑖𝑛 + 1 𝑉 ≥ 1.1 𝑉
𝑉𝐸 = 2 𝑉
𝑅𝐸 =
𝑉𝐸
𝐼𝐸
=
2 𝑉
2.80 𝑚𝐴
= 714.29 𝛺
𝑅𝐸
′′ = 𝑅𝐸 − 𝑅𝐸
′ = 714.29 𝛺 − 30 𝛺 = 684.29 Ω
↑ 750 Ω
↓ 680 Ω
𝑹𝑬
′′ = 𝟔𝟖𝟎 𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐸 = 𝐼𝐸 × 𝑅𝐸 = 𝐼𝐸 × (𝑅𝐸
′ + 𝑅𝐸
′′) = (2.80 𝑚𝐴)(30 𝛺 + 680 𝛺) = 1.99 𝑉
𝑉𝐵 = 𝑉𝐸 + 𝑉𝐵𝐸 = 1.99 𝑉 + 0.7 𝑉 = 2.69 𝑉
𝐼2 = 10 × 𝐼𝐵 = 10(56 µ𝐴) = 0.56 𝑚𝐴
𝑅𝐵2 =
𝑉𝐵
𝐼2
=
2.69 𝑉
0.56 𝑚𝐴
= 4.80 𝑘𝛺
↑ 5.1 𝑘Ω
↓ 4.7 𝑘Ω
𝑹𝑩𝟐 = 𝟓. 𝟏 𝒌𝜴 𝒎𝒆𝒋𝒐𝒓𝒂 𝒁𝒊𝒏
𝑉𝐶𝐶 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 + 𝑉𝐸 = 8.41 𝑉 + 6.5 𝑉 + 1.99 𝑉 = 16.9 𝑉
𝑽𝑪𝑪 = 𝟏𝟖 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝑉𝑅𝐵1 = 𝑉𝐶𝐶 − 𝑉𝐵 = 18 𝑉 − 2.69 𝑉 = 15.31 𝑉
𝐼1 = 𝐼2 + 𝐼𝐵 = 0.56 𝑚𝐴 + 56 µ𝐴 = 0.62 𝑚𝐴
𝑅𝐵1 =
𝑉𝑅𝐵1
𝐼1
=
15.31 𝑉
0.62 𝑚𝐴
= 24.69 𝑘𝛺
↑ 27 𝑘Ω
↓ 24 𝑘Ω
𝑹𝑩𝟏 = 𝟐𝟒 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
|𝐴𝑉| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ =
1.57 𝑘𝛺
9.29 𝛺 + 30 𝛺
= 39.96
𝑍𝑖𝑛 = 𝑅𝐵1 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛−𝑇 = 24 𝑘Ω ∥ 5.1 𝑘Ω ∥ 2.0 𝑘Ω = 1.36 𝑘Ω
𝐶𝐶 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.57 𝑘Ω
≥ 1.01 µ𝐹
𝑪𝑪 = 𝟐. 𝟐 µ𝑭
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸
′ )
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × (9.29 𝛺 + 30 Ω)
= 40.51µ𝐹
𝑪𝑬 = 𝟒𝟕µ𝑭
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.36 𝑘Ω
= 1.17 µ𝐹
𝑪𝑩 = 𝟐. 𝟐 µ𝑭
𝑅𝐵𝐵 𝑅𝐵1 𝑅𝐵2
𝑍𝑖𝑛
𝑅𝐶 = 3 𝑘𝛺
�̂�𝑜 = 4 𝑉
𝑅𝐿 = 3.3 𝑘𝛺
𝑅𝐿
′ = 1.57 𝑘Ω
𝑉𝑅𝐶 = 8.41 𝑉
𝐼𝐶 = 2.80 𝑚𝐴 = 𝐼𝐸
𝐼𝐵 = 56 µ𝐴
𝑟𝑒 = 9.29 Ω
𝑅𝐸
′ = 30 Ω
𝑉𝐶𝐸 =6.5 𝑉
𝑍𝑖𝑛−𝑇 = 2.0 𝑘𝛺
Aplicando el método 3:
𝑍𝑖𝑛 = 𝑅𝐵𝐵 ∥ 𝑍𝑖𝑛−𝑇 ≥ 1 𝑘Ω
𝑅𝐵𝐵 =
𝑍𝑖𝑛 × 𝑍𝑖𝑛−𝑇
𝑍𝑖𝑛−𝑇 − 𝑍𝑖𝑛
=
(1 𝑘Ω)(2.0 𝑘Ω)
2.0 𝑘Ω − 1 𝑘Ω
= 2.0 𝑘Ω
𝑅𝐵𝐵 = 𝑅𝐵1 ∥ 𝑅𝐵2
𝑅𝐵1 = 𝑅𝐵2
2𝑅𝐵𝐵 = 𝑅𝐵1 = 𝑅𝐵2 = 2(2.0 𝑘Ω) = 4.0 𝑘𝛺
↑ 4.3 𝑘Ω
↓ 3.9 𝑘Ω
𝑹𝑩𝟏 = 𝑹𝑩𝟐 = 𝟏𝟎 𝒌𝜴 𝒎𝒆𝒋𝒐𝒓𝒂 𝒁𝒊𝒏
𝑉𝐶𝐸 = �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛
𝑉𝐵 =
𝑉𝐶𝐶
2
𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 =
𝑉𝐶𝐶
2
− 0.7 𝑉
𝑉𝐶𝐶 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 + 𝑉𝐸 = 𝑉𝑅𝐶 + �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 +
𝑉𝐶𝐶
2
− 0.7 𝑉
𝑉𝐶𝐶 = 2(𝑉𝑅𝐶 + �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 − 0.7 𝑉)
𝑉𝐶𝐶 = 2(8.41 𝑉 + 4 𝑉 + 0.1 𝑉 + 2 𝑉 − 0.7 𝑉) = 27.62 𝑉
𝑽𝑪𝑪 = 𝟑𝟎 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝑉𝐵 =
𝑉𝐶𝐶
2
=
30 𝑉
2
= 15 𝑉
𝑉𝐸 =
𝑉𝐶𝐶
2
− 0.7 𝑉 =
30 𝑉
2
− 0.7 𝑉 = 14.3 𝑉
𝑅𝐸 =
𝑉𝐸
𝐼𝐸
=
14.3 𝑉
2.80 𝑚𝐴
= 5.11 𝑘Ω
𝑅𝐸
′′ = 𝑅𝐸 − 𝑅𝐸
′ = 5.11 𝑘𝛺 − 30 𝛺 = 5.08 𝑘Ω
↑ 5.1 𝑘Ω
↓ 4.7 𝑘Ω
𝑹𝑬
′′ = 𝟓. 𝟏 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
|𝐴𝑉| =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′ =
1.57 𝑘Ω
9.29 𝛺 + 30 𝛺
= 39.96
𝑍𝑖𝑛 = 𝑅𝐵1 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛−𝑇 = 10 𝑘Ω ∥ 10 𝑘Ω ∥ 2.0 𝑘Ω = 1.43 𝑘Ω
𝐶𝐶 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.57 𝑘Ω
≥ 1.01 µ𝐹
𝑪𝑪 = 𝟐. 𝟐µ𝑭
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸
′ )
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × (9.29 𝛺 + 30 Ω)
= 40.51µ𝐹
𝑪𝑬 = 𝟒𝟕 µ𝑭
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 1.43 𝑘Ω
= 1.11 µ𝐹
𝑪𝑩 = 𝟐. 𝟐 µ𝑭
𝐴𝑖 =
𝑖𝑂
𝑖𝑖
=
𝑖𝑐
𝑖𝑒
= 𝛼 [1.25]
𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
= −
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵
(𝛽 + 1)
[1.26]
𝑅𝐵𝐵 = 𝑅𝐵1 ‖ 𝑅𝐵2
𝐴𝑣 ≈ −
𝑅𝐿
′
𝑍𝑖𝑛𝑇
[1.27]
𝑍𝑖𝑛𝑇 = 𝑟𝑒 +
𝑅𝐵𝐵
(𝛽 + 1)
[1.28]
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇||𝑅𝐸 [1.29]
𝑍𝑜𝑇 >
1
ℎ𝑜𝑒
[1.30]
𝑍𝑜 = 𝑍𝑜𝑇||𝑅𝐿
′
𝐶𝑜𝑚𝑜 𝑍𝑜𝑇 ≫ 𝑅𝐿
′
𝑍𝑜 = 𝑅𝐿
′ [1.31]
𝐂𝐁
𝐴𝑖 =
𝑖𝑂
𝑖𝑖
=
𝑖𝑐
𝑖𝑒
= 𝛼 [1.32]
𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
= −
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵||𝑋𝐶𝐵
(𝛽 + 1)
[1.33]
𝑆𝑖 𝑅𝐵 ≫ 𝑋𝐶𝐵
𝑅𝐵||𝑋𝐶𝐵 ≈ 𝑋𝐶𝐵
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒 +
𝑋𝐶𝐵
(𝛽 + 1)
𝑟𝑒 ≫
𝑋𝐶𝐵
(𝛽 + 1)
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒
[1.34]
𝑍𝑖𝑛𝑇 = 𝑟𝑒 [1.35]
𝐶𝑜𝑚𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝐶𝐵 → 𝑅𝐵𝐵 𝑠𝑒𝑟á 𝑖𝑔𝑢𝑎𝑙 𝑎 𝑐𝑒𝑟𝑜
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇||𝑅𝐸 = 𝑟𝑒 𝑅𝐸 [1.36]
𝑋𝐶𝐶 ≪ 𝑅𝐿
′
𝐶𝐶 ≥
10
𝜔𝑅𝐿
′
[1.37]
𝑋𝐶𝐸 ≪ 𝑍𝑖𝑛
𝐶𝐸 ≥
10
𝜔𝑍𝑖𝑛
[1.38]
𝐶𝑜𝑛: 𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
=
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵||𝑋𝐶𝐵
(𝛽 + 1)
𝑆𝑖 𝑅𝐵 ≫ 𝑋𝐶𝐵
𝑅𝐵𝐵||𝑋𝐶𝐵 ≈ 𝑋𝐶𝐵
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 +
𝑋𝐶𝐵
(𝛽 + 1)
𝑟𝑒 ≫
𝑋𝐶𝐵
(𝛽 + 1)
𝑋𝐶𝐵 ≪ 𝑟𝑒(𝛽 + 1)
𝐶𝐵 ≥
10
𝜔 ∗ 𝑟𝑒 ∗ (𝛽 + 1)
[1.39]
ω = 2πfmin)
𝐴𝑣 = 20
𝛽𝑚𝑖𝑛 = 100
𝑉𝑖𝑛 = 0.1 𝑆𝑒𝑛(𝜔𝑡) 𝑉
𝑓𝑚𝑖𝑛 = 1 𝑘𝐻𝑧
𝑅𝐿 = 1 𝑘𝛺
𝑍𝑖𝑛 ≥ 100 𝛺
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵
𝛽 + 1
=
𝑅𝐿
′
𝑍𝑖𝑛𝑇
𝑍𝑖𝑛 = 𝑅𝐸 ∥ 𝑍𝑖𝑛𝑇
𝑅𝐸 ≫ 𝑍𝑖𝑛𝑇
𝑍𝑖𝑛 ≈ 𝑍𝑖𝑛𝑇
𝐴𝑣 =
𝑅𝐿
′
𝑍𝑖𝑛
𝑍𝑖𝑛𝑚á𝑥
=
𝑅𝐿
′
𝑚á𝑥
𝐴𝑣
=
𝑅𝐿
𝐴𝑣
¿ 𝐸𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑖𝑠𝑒ñ𝑎𝑟 𝑐𝑜𝑛 𝑒𝑠𝑡𝑎𝑠 𝑐𝑜𝑛𝑑𝑖𝑐𝑖𝑜𝑛𝑒𝑠?
𝑍𝑖𝑛𝑚á𝑥
=
𝑅𝐿
𝐴𝑣
=
1 𝑘𝛺
20
= 50 𝛺
𝑁𝑜 𝑒𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑖𝑠𝑒ñ𝑎𝑟
Por lo anterior se modifica el problema:
𝑅𝐿 = 6.8 𝑘𝛺
𝑍𝑖𝑛𝑚á𝑥
=
𝑅𝐿
𝐴𝑣
=
6.8 𝑘𝛺
20
= 340 𝛺
𝑆𝑖 𝑒𝑠 𝑝𝑜𝑠𝑖𝑏𝑙𝑒 𝑑𝑖𝑠𝑒ñ𝑎𝑟
100 𝛺 < 𝑍𝑖𝑛 < 340 𝛺
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑍𝑖𝑛 = 200 𝛺
𝑅𝐿
′ ≥ 𝐴𝑣 × 𝑍𝑖𝑛 ≥ (20)(200 𝛺) = 4.0 𝑘𝛺
𝑅𝐶 ≥
𝑅𝐿 × 𝑅𝐿
′
𝑅𝐿 − 𝑅𝐿
′ ≥
6.8 𝑘Ω × 4.0 𝑘𝛺
6.8 𝑘𝛺 − 4.0 𝑘Ω
≥ 9.71 𝑘Ω
↑ 10 𝑘Ω
↓ 9.1 𝑘Ω
𝑹𝑪 = 𝟏𝟎 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
�̂�𝑜 = |𝐴𝑣| × �̂�𝑖𝑛 = 20 × 0.1 𝑉 = 2 𝑉
𝑅𝐿 = 6.8 𝑘𝛺
𝑅𝐿
′ = 𝑅𝐶 ∥ 𝑅𝐿 = 10 𝑘Ω ∥ 6.8 𝑘Ω = 4.05 𝑘Ω
𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅𝐿
′ × �̂�𝑜 ≥
10 𝑘𝛺
4.05 𝑘Ω
(2 𝑉) ≥ 4.4 𝑉
𝑽𝑹𝑪 = 𝟏𝟏 𝑽
𝐼𝐶 =
𝑉𝑅𝐶
𝑅𝐶
=
11 𝑉
10 𝑘Ω
= 1.1 𝑚𝐴 = 𝐼𝐸
1𝑚𝐴 < 𝐼𝐶 < 10 𝑚𝐴
𝐿𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝐼𝐶 𝑐𝑢𝑚𝑝𝑙𝑒 𝑐𝑜𝑛 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑒𝑙 𝑇𝐵𝐽.
𝐼𝐵 =
𝐼𝐶
𝛽𝑚𝑖𝑛
=
1.1 𝑚𝐴
100
= 10.89 µ𝐴
𝑟𝑒 =
𝑉𝑇
𝐼𝐶
=
26 𝑚𝑉
1.1 𝑚𝐴
= 23.64 Ω
𝑉𝐶𝐸 ≥ �̂�𝑜 − �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 ≥ 2 𝑉 − 0.1 𝑉 + 2 𝑉 ≥ 3.9 𝑉
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵
𝛽 + 1
𝑅𝐵𝐵 = (
𝑅𝐿
′
𝐴𝑣
− 𝑟𝑒) (𝛽 + 1) = (
4.0 𝑘𝛺
20
− 23.64 𝛺) (100 + 1) = 17.81 𝑘𝛺
𝑉𝑅𝐵1 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 − 𝑉𝐸𝐵 = 11 𝑉 + 3.9 𝑉 − 0.7 𝑉 = 15.6 𝑉
𝐼1 = 11 × 𝐼𝐵 = 11(10.89 µ𝐴) = 0.12 𝑚𝐴
𝑅𝐵1 =
𝑉𝑅𝐵1
𝐼1
=
15.6 𝑉
0.12 𝑚𝐴
= 130 𝑘Ω
↑ 150 𝑘Ω
↓ 130 𝑘Ω
𝑹𝑩𝟏 = 𝟏𝟑𝟎 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑅𝐵2 >
𝑅𝐵𝐵 × 𝑅𝐵1
𝑅𝐵1 − 𝑅𝐵𝐵
>
(17.81 𝑘Ω)(130 𝑘Ω)
130 𝑘Ω − 17.81 𝑘Ω
> 20.64 𝑘Ω
↑ 22 𝑘Ω
↓ 20 𝑘Ω
𝑹𝑩𝟐 = 𝟐𝟎 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑅𝐵𝐵 = 𝑅𝐵1 ∥ 𝑅𝐵2 = 130 𝑘Ω ∥ 20 𝑘Ω = 17.33 𝑘Ω
𝑍𝑖𝑛𝑇 = 𝑟𝑒 +
𝑅𝐵𝐵
𝛽 + 1
= 23.64 Ω +
17.33 𝑘Ω
100 + 1
= 195.22 Ω
𝐼2 = 𝐼1 − 𝐼𝐵 = 0.12 𝑚𝐴 − 10.89 µ𝐴 = 0.11 𝑚𝐴
𝑆𝑒 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎: 𝑉𝐵 ≥ 𝑉𝐸 + 𝑉𝐵𝐸 ≥ 1 𝑉 + 0.7 ≥ 1.7 𝑉
𝑉𝐵 = 𝑅𝐵2 × 𝐼2 = (20 𝑘Ω)(0.11 𝑚𝐴) = 2.2 𝑉 ≥ 1.7 𝑉
𝑆í 𝑐𝑢𝑚𝑝𝑙𝑒
𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 2.2 𝑉 − 0.7 𝑉 = 1.5 𝑉
𝑆𝑒 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎: 𝑉𝐸 ≥ 𝑉𝑖𝑛 + 1 𝑉 ≥ 1.1 𝑉
𝑅𝐸 =
𝑉𝐸
𝐼𝐸
=
1.5 𝑉
1.1 𝑚𝐴
= 1.36 𝑘Ω
↑ 1.5Ω
↓ 1.3Ω
𝑹𝑬 = 𝟏. 𝟑 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑉𝐸 = 𝐼𝐸 × 𝑅𝐸 = (1.1 𝑚𝐴)(1.3 𝑘𝛺) = 1.43 𝑉
𝑉𝐶𝐶 = 𝑉𝑅𝐶 + 𝑉𝐶𝐸 + 𝑉𝐸 = 11 𝑉 + 3.9 𝑉 + 1.43 𝑉 = 16.33 𝑉
𝑽𝑪𝑪 = 𝟏𝟖 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝑍𝑖𝑛 = 𝑅𝐸 ∥ 𝑍𝑖𝑛𝑇 = 1.3 𝑘𝛺 ∥ 195.22 Ω = 169.73Ω
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 +
𝑅𝐵𝐵
𝛽 + 1
=
4 kΩ
23.64 Ω +
17.33 𝑘Ω
100 + 1
= 20.48
𝐶𝐶 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 kHz × 4.0 kΩ
≥ 0.39 µF
𝑪𝑪 = 𝟏 µF
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 169.73 Ω
= 9.38 µ𝐹
𝑪𝑬 = 𝟏𝟎 µF
𝑅𝐶𝐶
𝑅𝐶 𝑉𝑐𝑐
𝐴𝑖 =
𝑖𝐸
𝑖𝐵
= 𝛽 + 1 [1.40]
𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
≈ −
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐿
′
𝐴𝑣 ≈ 1 [1.41]
𝑍𝑖𝑛𝑇 =
𝑉𝑖𝑛𝑇
𝑖𝑖𝑛𝑇
= (𝛽 + 1)(𝑟𝑒 + 𝑅𝐿
′ ) [1.42]
𝑍𝑖𝑛=𝑍𝑖𝑛𝑇||𝑅𝐵𝐵 [1.43]
𝑍𝑜𝑇 =
𝑉𝑜𝑇
𝑖𝑜𝑇
= 𝑟𝑒 +
𝑅𝐵𝐵
𝛽 + 1
[1.44]
𝑍𝑜 = 𝑍𝑜𝑇||𝑅𝐿
′
𝐶𝑜𝑚𝑜 𝑍𝑜𝑇 ≪ 𝑅𝐿
′
𝑍𝑜 ≈ 𝑍𝑜𝑇 [1.45]
𝑋𝐶𝐵 ≪ 𝑍𝑖𝑛 → 𝐶𝐵 ≥
10
𝜔𝑍𝑖𝑛
[1.46]
𝑋𝐶𝐸 ≪ 𝑅𝐿 → 𝐶𝐸 ≥
10
𝜔𝑅𝐿
[1.47]
𝐸𝑛 𝑒𝑙 𝑐𝑎𝑠𝑜 𝑑𝑒 𝑢𝑠𝑎𝑟 𝑅𝐶 ∶ 𝐶𝐶 =
10
𝜔𝑅𝐶
[1.48]
𝛽𝑚𝑖𝑛 = 80
𝑉𝑖𝑛 = 2 𝑆𝑒𝑛(𝜔𝑡) 𝑉
𝑓𝑚𝑖𝑛 = 1 𝑘𝐻𝑧
𝑅𝐿 = 470 kΩ
𝑍𝑖𝑛 ≥ 2.0 kΩ
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑚á𝑥𝑖𝑚𝑎 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐸
′
(𝑟𝑒 + 𝑅𝐸
′ )𝑚á𝑥 =
𝑅𝐿
′
𝑚á𝑥
𝐴𝑣
=
𝑅𝐿
𝐴𝑣
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛𝑚í𝑛𝑖𝑚𝑎 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜
𝑍𝑖𝑛 = 𝑅𝐵𝐵 ∥ 𝑍𝑖𝑛𝑇
𝑍𝑖𝑛𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐸
′ )
𝑆𝑖 𝑅𝐵𝐵 ≫ 𝑍𝑖𝑛𝑇
(𝑟𝑒 + 𝑅𝐸
′ )𝑚𝑖𝑛 =
𝑍𝑖𝑛𝑇𝑚𝑖𝑛
(𝛽 + 1)
=
𝑍𝑖𝑛
(𝛽 + 1)
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑎 𝑑𝑖𝑠𝑒ñ𝑜
𝑍𝑖𝑛
𝛽 + 1
< 𝑟𝑒 + 𝑅𝐸
′ <
𝑅𝐿
|𝐴𝑣|
𝑍𝑖𝑛𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐿
′ )
𝑍𝑖𝑛 = 𝑅𝐵𝐵 ∥ 𝑍𝑖𝑛𝑇 ≈ 𝑍𝑖𝑛𝑇 ≥ 2.0 kΩ
𝑆𝑖 𝑟𝑒 ≪ 𝑅𝐿
′
𝑅𝐿
′ >
𝑍𝑖𝑛
(𝛽 + 1)
>
2.0 𝑘𝛺
100 + 1
= 19.8 Ω
𝑅𝐸 >
𝑅𝐿 × 𝑅𝐿
′
𝑅𝐿 − 𝑅𝐿
′ >
470 Ω × 19.8 Ω
470 Ω − 19.8 Ω
> 20.67 Ω
↑ 22 Ω
↓ 20 Ω
𝑹𝑬 = 𝟏 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑅𝐿
′ = 𝑅𝐸 ∥ 𝑅𝐿 = 1 kΩ ∥ 470 Ω = 319.73 Ω
�̂�𝑜 = �̂�𝑖𝑛 = 2 𝑉
𝑉𝐸 =
𝑅𝐸
𝑅𝐿
′ × �̂�𝑜 =
1 kΩ
319.73 Ω
(2 V)(1.2) = 7.5 V
1.2 = 𝐹𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
𝑉𝐶𝐶 = 2 × 𝑉𝐸 = 2(7.5 𝑉)
𝑽𝑪𝑪 = 𝟏𝟓 𝑽
𝐼𝐸 =
𝑉𝐸
𝑅𝐸
=
7.5 𝑉
1 𝑘Ω
= 7.5 𝑚𝐴
1 𝑚𝐴 < 𝐼𝐸 < 10 𝑚𝐴
𝐿𝑎 𝑐𝑜𝑟𝑟𝑖𝑒𝑛𝑡𝑒 𝐼𝐸 𝑐𝑢𝑚𝑝𝑙𝑒 𝑐𝑜𝑛 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑒𝑙 𝑇𝐵𝐽.
𝐼𝐵 =
𝐼𝐸
𝛽𝑚𝑖𝑛
=
7.5 𝑚𝐴
80
= 93.75 µ𝐴
𝑟𝑒 =
𝑉𝑇
𝐼𝐸
=
26 𝑚𝑉
7.5 𝑚𝐴
= 3.47 Ω
𝑉𝑅𝐵2 = 𝑉𝐵 = 𝑉𝐸 + 𝑉𝐵𝐸 = 7.5 𝑉 + 0.7 𝑉 = 8.2 𝑉
𝐼2 = 10 × 𝐼𝐵 = 10(93.75 µ𝐴) = 0.94 𝑚𝐴
𝑅𝐵2 =
𝑉𝑅𝐵2
𝐼2
=
8.2 𝑉
0.94 𝑚𝐴
= 8.72 𝑘Ω
↑ 9.1 𝑘Ω
↓ 8.2 𝑘Ω
𝑹𝑩𝟐 = 𝟗. 𝟏 𝐤𝛀 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝐼1 = 11 × 𝐼𝐵 = 11(93.75 µ𝐴) = 1.03 𝑚𝐴
𝑉𝑅𝐵1 = 𝑉𝐶𝐶 − 𝑉𝐵 = 15 𝑉 − 8.2 𝑉 = 6.8 𝑉
𝑅𝐵1 =
𝑉𝑅𝐵1
𝐼1
=
6.8 𝑉
1.03 𝑚𝐴
= 6.6 𝑘Ω
↑ 6.8 𝑘Ω
↓ 6.2 𝑘Ω
𝑹𝑩𝟏 = 𝟔. 𝟖 𝒌𝜴 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂.
𝑍𝑖𝑛−𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐿
′ ) = (80 + 1)(3.47 Ω + 319.73 Ω) = 26.18 kΩ
𝑅𝐵𝐵 = 𝑅𝐵1 ∥ 𝑅𝐵2 = 6.8 kΩ ∥ 9.1 kΩ = 3.89 kΩ
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒 + 𝑅𝐿
′ =
319.73 Ω
3.47 Ω + 319.73 Ω
= 0.99
𝑍𝑖𝑛 = 𝑅𝐵1 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛−𝑇 = 6.8 𝑘Ω ∥ 9.1 𝑘Ω ∥ 26.18 𝑘Ω = 3.39 𝑘Ω
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 319.73 Ω
= 4.97 µ𝐹
𝑪𝑬 = 𝟏𝟎 µ𝑭
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 3.39 𝑘Ω
= 0.47 µ𝐹
𝑪𝑩 = 𝟏 µ𝑭
t
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][5.4][1
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kRkHzf
ßVVkZ
Lmín
mínOin
𝐼𝐷
𝑉𝐺𝑆 𝑉𝐷𝑆
𝑉𝐺𝑆
𝐼𝐷
𝑉𝐺𝑆
𝐼𝐷 = 0 𝐴
𝑉𝐺𝑆 𝑉𝑃 𝐼𝐷𝑆𝑆 𝑉𝑃
𝑉𝑃 = 𝑉𝐺𝑆 𝑠𝑖 𝐼𝐷 = 0 𝐴
𝐼𝐷 = 𝐼𝐷𝑆𝑆 𝑠𝑖 𝑉𝐺𝑆 = 0 𝑉
𝑉𝑃 = 𝑉𝑝𝑖𝑛𝑐ℎ 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑅𝑐𝑎𝑛𝑎𝑙 → ∞
𝐼𝐷 = 𝐼𝐷𝑆𝑆 (1 −
𝑉𝐺𝑆
𝑉𝑝
)
2
[1.49]
𝑟𝑑: 𝑟𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑖𝑎 𝑑𝑖𝑛á𝑚𝑖𝑐𝑎, 𝑝𝑎𝑟á𝑚𝑒𝑡𝑟𝑜 𝑑𝑒𝑙 𝑓𝑎𝑏𝑟𝑖𝑐𝑎𝑛𝑡𝑒, [Ω]
𝑟𝑑 =
∆𝑉𝐷𝑆
∆𝐼𝐷𝑆
|
𝑉𝐺𝑆=𝑐𝑡𝑒
[1.50]
𝑔𝑚: 𝑡𝑟𝑎𝑛𝑠𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎, [Siems]
𝑔𝑚 ≅
∆𝐼𝐷
∆𝑉𝐺𝑆
|
𝑉𝐷𝑆=𝑐𝑡𝑒
[1.51]
𝑔𝑚
𝑔𝑚 = 𝑔𝑚𝑜 (1 −
𝑉𝐺𝑆
𝑉𝑃
) [1.52]
𝑔𝑚𝑜: 𝑡𝑟𝑎𝑛𝑠𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑖𝑎 𝑚á𝑥𝑖𝑚𝑎, 𝑚á𝑥𝑖𝑚𝑎 𝑝𝑒𝑛𝑑𝑖𝑒𝑛𝑡𝑒 𝑑𝑒 𝑙𝑎 𝑐𝑢𝑟𝑣𝑎.
𝑔𝑚𝑜 = −
2𝐼𝐷𝑆𝑆
𝑉𝑃
[1.53]
𝑔𝑚 = −
2𝐼𝐷𝑆𝑆
𝑉𝑃
(1 −
𝑉𝐺𝑆
𝑉𝑃
) [1.54]
𝑔𝑚 =
𝜕𝑖𝐷
𝜕𝑉𝐺𝑠
≈
∆𝐼𝐷
∆𝑉𝐺𝑆
|
𝑉𝐷𝑆=𝑐𝑡𝑒
𝑔𝑑 =
1
𝑟𝑑
=
𝜕𝑖𝐷
𝜕𝑉𝐷𝑠
≈
∆𝐼𝐷
∆𝑉𝐷𝑆
|
𝑉𝐷𝑆=𝑐𝑡𝑒
𝜇: 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑐𝑖ó𝑛
𝜇 = 𝑔𝑚(𝑟𝑑) [1.55]
𝑟𝑑 = 100 kΩ
𝑔𝑚 = 4.45 mS
→ 𝜇 = 445
𝐴𝑣 =
−𝜇𝑅𝐿
′
𝑟𝑑 + 𝑅𝐿
′ + (𝜇 + 1)𝑅𝑠
′
[1.56]
𝑠𝑖 𝑟𝑑 ≫ 𝑅𝐿
′ + (𝜇 + 1)𝑅𝑠
′ [1.57]
𝐴𝑣𝑚𝑎𝑥 =
−𝜇𝑅𝐿
′
𝑟𝑑
= −𝑔𝑚𝑅𝐿
′ [1.58]
𝑉𝑅𝐷
≥
𝑅𝐷
𝑅𝐿
′ �̂�𝑜 [1.59]
𝑉𝐷𝑆 > �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐷𝑆𝑚𝑖𝑛 [1.60]
𝑉𝐷𝑆𝑚𝑖𝑛 > |𝑉𝑝| [1.61]
𝑉𝑠 > 𝑉𝑖𝑛 [1.62]
𝑉𝐶𝐶 > 𝑉𝑆 + �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐷𝑆𝑚𝑖𝑛 + 𝑉𝑅𝐷 [1.63]
𝑍𝑖𝑛𝑇 = ∞ [1.64]
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇||𝑅𝐺 = 𝑅𝐺 [1.65]
𝑍𝑜𝑇 = 𝑟𝑑 [1.66]
𝑍𝑜 = 𝑅𝐿||𝑟𝐷 = 𝑅𝐿
′ [1.67]
|𝐴𝑣| = 10
𝑉𝑖𝑛 = 0.1 𝑆𝑒𝑛(𝜔𝑡) 𝑉
𝑓𝑚𝑖𝑛 = 1 𝑘𝐻𝑧
𝑅𝐿 = 6.8 kΩ
𝑍𝑖𝑛 ≥ 100 𝑘Ω
𝐷𝑎𝑡𝑜𝑠 𝑑𝑒𝑙 𝐽𝐹𝐸𝑇
𝐼𝐷𝑆𝑆 = 10 𝑚𝐴
𝑉𝑝 = −4 𝑉
𝑟𝑑 = 120 𝑘Ω
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑚á𝑥𝑖𝑚𝑎 𝑑𝑒 𝑙𝑎 𝑔𝑎𝑛𝑎𝑐𝑖𝑎
𝐴𝑣𝑚á𝑥
= −𝑔𝑚 × 𝑅𝐿
′
|𝐴𝑣𝑚á𝑥| = 𝑔𝑚 × 𝑅𝐿
′
𝑔𝑚𝑜 = −2
𝐼𝐷𝑆𝑆
𝑉𝑝
= −2
10 𝑚𝐴
−4
= 5 𝑚𝑆
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑹𝑫 = 𝟔. 𝟖 𝒌𝛀
�̂�𝑜 = |𝐴𝑣| × �̂�𝑖𝑛 = 10 × 0.1 𝑉 = 1 𝑉
𝑅𝐿 = 6.8 𝑘𝛺
𝑅𝐿
′ = 𝑅𝐷 ∥ 𝑅𝐿 = 6.8 kΩ ∥ 6.8 kΩ = 3.4 kΩ
𝑉𝑅𝐷 =
𝑅𝐷
𝑅𝐿
′ × �̂�𝑜 =
6.8 kΩ
3.4 kΩ
(1 V) = 2 V × 1.2 = 2.4 V
1.2 = 𝐹𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
𝐼𝐷𝑆𝑆 = 10 𝑚𝐴
1
3
𝐼𝐷𝑆𝑆 < 𝐼𝐷 <
2
3
𝐼𝐷𝑆𝑆
1.33 𝑚𝐴 < 𝐼𝐷 < 6.67 𝑚𝐴
𝐼𝐷 =
𝑉𝑅𝐷
𝑅𝐷
=
2.4 𝑉
6.8 𝑘Ω
= 0.35 𝑚𝐴 𝑛𝑜 𝑐𝑢𝑚𝑝𝑙𝑒 𝑙𝑎 𝑐𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑉𝑅𝐷 = 15 𝑉
𝐼𝐷 =
𝑉𝑅𝐷
𝑅𝐷
=
15 𝑉
6.8 kΩ
= 2.21 𝑚𝐴 = 𝐼𝑆
𝐼𝐷 = 𝐼𝐷𝑆𝑆 (1 −
𝑉𝐺𝑆
𝑉𝑃
)
2
𝑉𝐺𝑆 = (1 − √
𝐼𝐷
𝐼𝐷𝑆𝑆
) 𝑉𝑃 = (1 − √
2.21 𝑚𝐴
10 𝑚𝐴
) (−4 𝑉) = −2.12 𝑉
𝑔𝑚 = −2
𝐼𝐷𝑆𝑆
𝑉𝑝
(1 −
𝑉𝐺𝑆
𝑉𝑃
) = −2
10 𝑚𝐴
−4 𝑉
(1 −
−2.12 𝑉
−4 𝑉
) = 2.35 𝑚𝑆
|𝐴𝑣|𝑚𝑎𝑥 = 𝑔𝑚 × 𝑅𝐿
′ = 2.35 𝑚𝑆 × 3.4 kΩ = 7.99
⇒ 𝑛𝑜 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑐𝑢𝑚𝑝𝑙𝑖𝑟 𝑐𝑜𝑛 𝑙𝑎 𝑔𝑎𝑛𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 10
𝐴𝑠𝑢𝑚𝑖𝑟: |𝐴𝑣| = 7.5
𝐴𝑣 =
−𝜇𝑅𝐿
′
𝑟𝑑 + 𝑅𝐿
′ + (𝜇 + 1)𝑅𝑠
′
𝜇 = 𝑔𝑚 × 𝑟𝑑 = (2.35 mS)(120 kΩ) = 282
𝑅𝑠
′ = (
−𝜇𝑅𝐿
′
𝐴𝑣
− 𝑟𝑑 − 𝑅𝐿
′ )
1
(𝜇 + 1)
𝑅𝑠
′ = (
−(282)(3.4 kΩ)
−7.5
− 120 kΩ − 3.4 kΩ)
1
(282 + 1)
= 15.69 Ω
↑ 16 Ω
↓ 15 Ω
𝑹𝒔
′ = 𝟏𝟓 𝛀 𝒑𝒂𝒓𝒂 𝒎𝒂𝒏𝒕𝒆𝒏𝒆𝒓 𝒈𝒂𝒏𝒂𝒏𝒄𝒊𝒂
𝑉𝐺𝑆 = −𝑉𝑆 ⇒ 𝑉𝑆 = 2.12 V
𝑅𝑠 =
𝑉𝑠
𝐼𝑆
=
2.12 V
2.21 mA
= 959.28 Ω
𝑅𝑠
′′ = 𝑅𝑠 − 𝑅𝑠
′ = 959.28 Ω − 15 Ω = 944.28 Ω
↑ 1 kΩ
↓ 910 Ω
𝑹´´𝒔 = 𝟏 𝐤𝛀 𝒑𝒐𝒓 𝒄𝒆𝒓𝒄𝒂𝒏í𝒂
𝑅𝑠 = 𝑅𝑠
′ + 𝑅𝑠
′′ = 15 Ω + 1 kΩ = 1.015 kΩ
𝑉𝑅𝑆 = 𝐼𝑆𝑅𝑠 = (2.21 𝑚𝐴)(1.015 kΩ) = 2.24 𝑉
𝑉𝐷𝑆 > |𝑉𝑃| ⇒ 𝑉𝐷𝑆 = 4 𝑉
�̂�𝑜 = |𝐴𝑣| × �̂�𝑖𝑛 = 7.5 × 0.1 𝑉 = 0.75 𝑉
𝑉𝐷𝐷 ≥ 𝑉𝑅𝐷 + �̂�𝑜 + �̂�𝑖 + 𝑉𝐷𝑆𝑚𝑖𝑛 + 𝑉𝑆 = 15 𝑉 + 0.75 𝑉 + 0.1 𝑉 + 4 𝑉 + 2.24 𝑉
= 22.09 𝑉
𝑽𝑫𝑫 = 𝟐𝟒 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝐴𝑣 =
−𝜇𝑅𝐿
′
𝑟𝑑 + 𝑅𝐿
′ + (𝜇 + 1)𝑅𝑠
′ =
−(282)(3.4 kΩ)
120 kΩ + 3.4 kΩ + (282 + 1)(15 Ω)
= −7.51
𝑍𝑖𝑛 = 𝑅𝐺 ≥ 100 𝑘𝛺 ⇒ 𝑹𝑮 = 𝟏𝟐𝟎 𝐤𝛀
𝑋𝐺𝐺 ≪ 𝑅𝐺
𝐶𝐺 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐺
𝐶𝐺 ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 120 𝑘Ω
≥ 0.013 𝜇𝐹
𝑪𝑮 = 𝟎. 𝟏µ𝑭
𝑋𝐺𝑆 ≪ 𝑅𝑆
𝐴𝑣 =
−𝜇𝑅𝐿
′
𝑟𝑑 + 𝑅𝐿
′ + (𝜇 + 1)𝑋𝐶𝑆𝑟𝑑 + 𝑅𝐿
′ ≫ 𝜇𝑋𝐶𝑆
𝜇 ≫ 1
∴ 𝜇 = 𝑔𝑚 × 𝑟𝑑
𝑟𝑑 ≫ 𝜇𝑋𝐶𝑆
𝑋𝐶𝑆 ≪
𝑟𝑑
𝜇
=
1
𝑔𝑚
𝐶𝑆 ≥
10 × 𝑔𝑚
2𝜋 × 𝑓𝑚𝑖𝑛
𝐶𝑆 ≥
10(2.35 𝑚𝑆)
2𝜋 × 1 𝑘𝐻𝑧
≥ 3.74 𝜇𝐹
𝑪𝑺 = 𝟒. 𝟕 𝝁𝑭
𝑋𝐶𝐷 ≪ 𝑅𝐷
′ + 𝑅𝐿
𝑅𝐷
′ = 𝑟𝑑 ∥ 𝑅𝐷
𝑟𝑑 ≫ 𝑅𝐷
𝑅𝐷
′ ≈ 𝑅𝐷
𝑋𝐶𝐷 ≪ 𝑅𝐷 + 𝑅𝐿
𝐶𝐷 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑅𝐷 + 𝑅𝐿)
𝐶𝐷 ≥
10
2𝜋 × 1 𝑘𝐻𝑧 × (6.8 kΩ + 6.8 kΩ)
≥ 0.12 𝜇𝐹
𝑪𝑫 = 𝟎. 𝟑𝟑 𝝁𝑭
𝑉𝑜 =
𝑉𝑖 × 𝑅𝐿
𝑅𝐷
′ + 𝑋𝐶𝐷 + 𝑅𝐿
𝑋𝐶𝐷 ≪ 𝑅𝐷
′ + 𝑅𝐿
𝑅𝐷
′ ≈ 𝑅𝐷
𝐶𝐷 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑅𝐷 + 𝑅𝐿
′ )
𝐶𝐷.
𝐴𝑣 =
𝑔𝑚
𝑔𝑚 + 𝑅𝐿
′ −1
+ 𝑔𝑑
[1.68]
𝑅𝐿
′ = 𝑅𝐿‖𝑅𝑆 𝑦 𝑔𝑑 =
1
𝑟𝑑
[1.69]
𝐴𝑣𝑚𝑎𝑥 =
𝑔𝑚
𝑔𝑚 + 𝑔𝑑
≈ 1 𝑅𝐿
′ 𝑚𝑢𝑦 𝑔𝑟𝑎𝑛𝑑𝑒 →
1
𝑅𝐿
′ ≈ 0 [1.70]
𝑔𝐺 =
1
𝑅𝐺
→ 𝑍𝑖 = 𝑅𝐺 [1.71]
𝑅𝑂 = 𝑅𝐿
′ ‖𝑟𝑑‖ (
1
𝑔𝑚
) [1.72]
𝑅𝑂 =
1
𝑔𝑚
[1.73]
𝑉𝑖𝑛 = 0.1 𝑆𝑒𝑛(𝜔𝑡) V
𝑓𝑚𝑖𝑛 = 1 kHz 𝑅𝐿 = 3 kΩ 𝑍𝑖𝑛 ≥ 10 MΩ
𝐷𝑎𝑡𝑜𝑠 𝑑𝑒𝑙 𝐽𝐹𝐸𝑇
𝐼𝐷𝑆𝑆 = 10 mA
𝑉𝑝 = −4 V
𝑟𝑑 = 120 kΩ
𝐴𝑣𝑚á𝑥
=
𝑔𝑚 × 𝑅𝐿
′
1 + 𝑔𝑚𝑅𝐿
′
𝑅𝐿
′ = 𝑅𝐷 ∥ 𝑅𝐿
𝑉𝑆 =
𝑅𝑆
𝑅𝐿
′ × �̂�𝑜
𝑅𝑆
𝑅𝐿
′ > 1
𝑅𝑆 > 𝑅𝐿
′ ≈ 𝑅𝐿
𝑅𝑆 > 3 kΩ
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒: 𝑹𝑺 = 𝟒. 𝟕 𝐤𝛀
𝑅𝐿
′ = 𝑅𝑆 ∥ 𝑅𝐿 = 4.7 kΩ ∥ 3 kΩ = 1.83 kΩ
𝑉𝑆 =
𝑅𝑆
𝑅𝐿
′ × �̂�𝑜 =
4.7 𝑘𝛺
1.83 𝑘𝛺
(2 𝑉) = 5.14 𝑉 × 1.2 = 6.16 𝑉
1.2 = 𝐹𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
𝐼𝐷𝑆𝑆 = 10 𝑚𝐴
1
3
𝐼𝐷𝑆𝑆 < 𝐼𝐷 <
2
3
𝐼𝐷𝑆𝑆
1.33 𝑚𝐴 < 𝐼𝐷 < 6.67 𝑚𝐴
𝑉𝑆 = 10 𝑉
𝐼𝑆 =
𝑉𝑆
𝑅𝑆
=
10 𝑉
4.7 𝑘𝛺
= 2.18 𝑚𝐴 = 𝐼𝐷
𝐼𝐷 = 𝐼𝐷𝑆𝑆 (1 −
𝑉𝐺𝑆
𝑉𝑃
)
2
𝑉𝐺𝑆 = (1 − √
𝐼𝐷
𝐼𝐷𝑆𝑆
) 𝑉𝑃 = (1 − √
2.18 𝑚𝐴
10 𝑚𝐴
) (−4 𝑉) = −2.13 𝑉
𝑔𝑚 = −2
𝐼𝐷𝑆𝑆
𝑉𝑝
(1 −
𝑉𝐺𝑆
𝑉𝑃
) = −2
10 𝑚𝐴
−4 𝑉
(1 −
−2.13 𝑉
−4 𝑉
) = 2.34 𝑚𝑆
𝑉𝐺𝐺 = 𝑉𝐺𝑆 + 𝑉𝑆 = −2.13 𝑉 + 10 𝑉 = 7.87 𝑉
𝑉𝐷𝐷 = 2 × 𝑉𝑆 = 2(10 𝑉) = 20 𝑉
𝑽𝑫𝑫 = 𝟐𝟏 𝑽 𝒆𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒊𝒛𝒂𝒅𝒐
𝐴𝑣 =
𝑔𝑚 × 𝑅𝐿
′
1 + 𝑔𝑚𝑅𝐿
′ =
(2.34 𝑚𝑆)(1.83 𝑘𝛺)
1 + (2.34 𝑚𝑆)(1.83 𝑘𝛺)
= 0.81
𝑍𝑖𝑛 = 𝑅𝐺 ≥ 10 MΩ ⇒ 𝑹𝑮 = 𝟏𝟎 𝐌𝛀
𝐶𝐺 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐺
≥
10
2𝜋 × 1 𝑘𝐻𝑧 × 10 𝑀Ω
≥ 0.16 µF
𝑪𝑮 = 𝟏 𝛍𝐅
𝐶𝑆 ≥
10 × 𝑔𝑚
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ =
10(2.34𝑚𝑆)
2𝜋 × 1𝐾𝐻𝑧 × 1.83𝐾𝛺
= 2.03 µF
𝑪𝑺 = 𝟏 𝛍𝐅
RL
′ = RD‖RL [1.74]
Av =
gm + gd
1
RL
′ + gd
≈ gmRL
′ [1.75]
Zin = RS‖
1
gm
[1.76]
Zo = RL
′ ‖rd ≈ RL
′ [1.77]
2011.0ˆ vLin AkRVV
kr
mAIVVpmSg
d
DSSm
90
4425.2
𝑉𝑐𝑐
𝑅𝑖𝑛 = 𝑟𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑒𝑛𝑡𝑟𝑎𝑑𝑎
𝑅𝑂 = 𝑟𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑠𝑎𝑙𝑖𝑑𝑎
𝐺𝑉𝑑 = 𝑓𝑢𝑒𝑛𝑡𝑒 𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑗𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑎𝑑𝑎
𝑉𝑑 = 𝑡𝑒𝑛𝑠𝑖ó𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑑𝑒 𝑒𝑛𝑡𝑟𝑎𝑑𝑎 𝑉+ − 𝑉−
𝐺 = 𝑔𝑎𝑛𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑗𝑒 𝑒𝑛 𝑙𝑎𝑧𝑜 𝑎𝑏𝑖𝑒𝑟𝑡𝑜
G
V-
V+
+ Vcc
- Vcc
VoIn
In Salida
𝑅𝑖𝑛 → ∞
𝑅𝑜 = 0 Ω
𝐺 → ∞
∆𝐵 → ∞
𝑉𝑜 = 0 V 𝐺𝑚𝑐 = 0 =
𝑉𝑜
𝑉𝑚𝑐
𝐶𝑅𝑀 =
𝐺𝑑
𝐺𝑚𝑐
∞
𝐺 → ∞
𝑉𝑜 = +𝐸 𝑠𝑖 𝑉𝑑 > 0
𝑉𝑜 = −𝐸 𝑠𝑖 𝑉𝑑 < 0
donde 𝐸 = 𝑣𝑜𝑙𝑡𝑎𝑗𝑒 𝑑𝑒 𝑠𝑎𝑙𝑖𝑑𝑎 𝑑𝑒 𝑠𝑎𝑡𝑢𝑟𝑎𝑐𝑖ó𝑛
𝐺
G
V-
V+
+ Vcc
- Vcc
Vo=G Vd=G (V+ - V-)
In
In
𝑉+ 𝑉− 𝑉+ = 𝑉−
𝑉+ 𝑉− 𝐼− = 𝐼+ = 0
(𝑉−)
(𝑉+)
𝑉+ = 𝑉−
𝑉− = 𝑉+
𝑉𝑑 → 0𝑉
𝐼+ = 𝐼− = 0𝐴
Ley de nodos de Kirchhoff
𝑽− :
𝑉𝑖−𝑉−
𝑅𝐴
+
𝑉𝑜−𝑉−
𝑅𝐹
= 0 V (𝒆𝒄. 𝟏)
𝑽+: 0 𝑉 − 𝑉+ = 0 𝑉 ⟹ 𝑉+ = 0 𝑉
𝑉− = 𝑉+
𝐸𝑐. 1:
𝑉𝑖 − 0 𝑉
𝑅𝐴
= −
𝑉𝑜 − 0 𝑉
𝑅𝐹
𝑉𝑖
𝑅𝐴
= −
𝑉𝑜
𝑅𝐹
𝑉𝑜
𝑉𝑖
= −
𝑅𝐴
𝑅𝐹
𝐺𝑎𝑛𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 𝑙𝑎𝑧𝑜 𝑐𝑒𝑟𝑟𝑎𝑑𝑜
𝑨𝒗 = −
𝑹𝑨
𝑹𝑭
[𝟏. 𝟕𝟖]
Función de transferencia es independiente de G y depende de las
resistencias externas.
La realimentación de la salida a la entrada a través de 𝑅𝐹 sirve para
llevar la tensión diferencial de 𝑉𝐷 = 𝑉𝑖𝑛 = 𝑉+ − 𝑉− a cero, es decir
𝑉𝐷 = 0 V.
Como 𝑉+ = 0 la realimentación negativa tiene el efecto de llevar a 𝑉− a
cero, por lo tanto en el A.O. V+ = V- = 0 y existe una tierra virtual. El
término virtual significa que V- = 0 (potencial de tierra) pero no fluye
corriente real en este cortocircuito, debido a que no puede fluir ninguna
corriente para V- y para V+ debido a Ren → .
𝑉− = 𝑉+
𝑉𝑑 → 0 V
𝐼+ = 𝐼− = 0 A
Ley de nodos de Kirchhoff
𝑽− :
𝑉𝐴 − 𝑉−
𝑅𝐴
+
𝑉𝐵 − 𝑉−
𝑅𝐵
+
𝑉𝐶 − 𝑉−
𝑅𝐶
+
𝑉𝑜 − 𝑉−
𝑅𝐹
= 0 V 𝑬𝒄. 𝟏
𝑽+: 0 − 𝑉+ = 0 ⟹ 𝑉+ = 0 V
𝑉− = 𝑉+
𝐸𝑐. 1:
𝑉𝐴 − 0
𝑅𝐴
+
𝑉𝐵 − 0
𝑅𝐵
+
𝑉𝐶 − 0
𝑅𝐶
+
𝑉𝑜 − 0
𝑅𝐹
= 0 V
𝑉𝐴
𝑅𝐴
+
𝑉𝐵
𝑅𝐵
+
𝑉𝐶
𝑅𝐶
+
𝑉𝑜
𝑅𝐹
= 0 V
𝑽𝒐 = −𝑹𝑭 (
𝑽𝑨
𝑹𝑨
+
𝑽𝑩
𝑹𝑩
+
𝑽𝑪
𝑹𝑪
)
𝑽𝒐 = −𝑹𝑭 ∑
𝑽𝒊
𝑹𝒊
𝒄
𝒊=𝒂
[𝟏. 𝟕𝟗]
𝑉− = 𝑉+
𝑉𝑑 → 0 V
𝐼+ = 𝐼− = 0 A
Ley de nodos de Kirchhoff
𝑽− :
0𝑉 − 𝑉−
𝑅𝐴
+
𝑉𝑜 − 𝑉−
𝑅𝐹
= 0 V 𝑬𝒄. 𝟏
𝑽+: 𝑉𝑖𝑛 − 𝑉+ = 0 V ⟹ 𝑉+ = 𝑉𝑖𝑛
𝑉− = 𝑉+
𝐸𝑐. 1:
0𝑉 − 𝑉𝑖𝑛
𝑅𝐴
+
𝑉𝑜 − 𝑉𝑖𝑛
𝑅𝐹
= 0 V
−𝑉𝑖𝑛
𝑅𝐴
+
𝑉𝑜 − 𝑉𝑖𝑛
𝑅𝐹
= 0 V
𝑉𝑖𝑛
𝑅𝐴
=
𝑉𝑜 − 𝑉𝑖𝑛
𝑅𝐹
𝑅𝐹
𝑅𝐴
=
𝑉𝑜−𝑉𝑖𝑛
𝑉𝑖𝑛
𝑅𝐹
𝑅𝐴
=
𝑉𝑜
𝑉𝑖𝑛
− 1
𝑽𝒐
𝑽𝒊𝒏
= 𝟏 +
𝑹𝑭
𝑹𝑨
𝑨𝒗 = 𝟏 +
𝑹𝑭
𝑹𝑨
[𝟏. 𝟖𝟎]
𝑉𝑖𝑛1 = −0.5 𝑆𝑒𝑛(𝜔1𝑡) V
𝑉𝑖𝑛2 = −1 𝑆𝑒𝑛(𝜔2𝑡) V
𝑅4 = 1 k𝛺
𝑅1 = 10 k𝛺
𝑉− = 𝑉+
𝑉𝑑 → 0 V
𝐼+ = 𝐼− = 0 A
𝑺𝒊: 𝑽𝒊𝒏𝟐 = 𝟎 𝐕
𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟 𝐴1
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽+: 𝑉𝑖𝑛1 − 𝑉+ = 0 V ⟹ 𝑉+ = 𝑉𝑖𝑛1
𝑽− :
0 𝑉 − 𝑉−
𝑎𝑅
+
𝑉𝑜1 − 𝑉−
𝑅
= 0 V
0𝑉 − 𝑉𝑖𝑛1
𝑎𝑅
+
𝑉𝑜1 − 𝑉𝑖𝑛1
𝑅
= 0 V
𝑉𝑜1 = 𝑉𝑖𝑛1 (1 +
𝑅
𝑎𝑅
) = 𝑉𝑖𝑛1 (1 +
1
𝑎
) 𝐸𝑐. 1
𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟 𝐴2
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽+: 0 V − 𝑉+ = 0 V ⟹ 𝑉+ = 0 V
𝑽− :
𝑉𝑖𝑛1 − 𝑉−
𝑎𝑅
+
𝑉𝑜2 − 𝑉−
𝑅
= 0 V
𝑉𝑖𝑛1 − 0 V
𝑎𝑅
+
𝑉𝑜2 − 0 V
𝑅
= 0 V
𝑉𝑜2 = −
𝑅
𝑎𝑅
𝑉𝑖𝑛1
𝑉𝑜2 = −
𝑉𝑖𝑛1
𝑎
𝐸𝑐. 2
𝑺𝒊: 𝑽𝒊𝒏𝟏 = 𝟎 𝐕
𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟 𝐴2
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽+: 𝑉𝑖𝑛2 − 𝑉+ = 0 V ⟹ 𝑉+ = 𝑉𝑖𝑛2
𝑽− :
0 V − 𝑉−
𝑎𝑅
+
𝑉𝑜2 − 𝑉−
𝑅
= 0 V
0 V − 𝑉𝑖𝑛2
𝑎𝑅
+
𝑉𝑜2 − 𝑉𝑖𝑛2
𝑅
= 0 V
𝑉𝑜2 = 𝑉𝑖𝑛2 (1 +
𝑅
𝑎𝑅
) = 𝑉𝑖𝑛2 (1 +
1
𝑎
) 𝐸𝑐. 3
𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟 𝐴1
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽+: 0𝑉 − 𝑉+ = 0 V ⟹ 𝑉+ = 0 V
𝑽− :
𝑉𝑖𝑛2 − 𝑉−
𝑎𝑅
+
𝑉𝑜1 − 𝑉−
𝑅
= 0 V
𝑉𝑖𝑛2 − 0 V
𝑎𝑅
+
𝑉𝑜1 − 0 V
𝑅
= 0 V
𝑉𝑜1 = −
𝑅
𝑎𝑅
𝑉𝑖𝑛2
𝑉𝑜2 = −
𝑉𝑖𝑛2
𝑎
𝐸𝑐. 4
𝐸𝑐. 1 + 𝐸𝑐. 4
𝑉𝑜1 = 𝑉𝑖𝑛1 (1 +
1
𝑎
) −
𝑉𝑖𝑛2
𝑎
𝐸𝑐. 5
𝐸𝑐. 2 + 𝐸𝑐. 3
𝑉𝑜2 = 𝑉𝑖𝑛2 (1 +
1
𝑎
) −
𝑉𝑖𝑛1
𝑎
𝐸𝑐. 6
𝐴𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟 𝐴3
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽− :
𝑉𝑖𝑛1 − 𝑉−
𝑅4
+
𝑉𝑜 − 𝑉−
𝑅1
= 0 V
𝑽+ : ∶
𝑉𝑜2 − 𝑉+
𝑅4
+
0 V − 𝑉+
𝑅1
= 0 V
𝑉𝑜2 − 𝑉+
𝑅4
+
−𝑉+
𝑅1
= 0 V
𝑉+ = 𝑉𝑜2
𝑅1
𝑅1 + 𝑅4
𝐸𝑐. 7
𝐸𝑐. 7 𝑟𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑟 𝑒𝑛 𝑉−
𝑉𝑖𝑛1 − 𝑉−
𝑅4+
𝑉𝑜 − 𝑉−
𝑅1
= 0 V
𝑉𝑜 =
𝑅1
𝑅4
(𝑉𝑜2 − 𝑉01) 𝐸𝑐. 8
𝐸𝑐. 5 𝑦 6 𝑒𝑛 𝐸𝑐. 8
𝑉𝑜 =
𝑅1
𝑅4
{[𝑉𝑖𝑛2 (1 +
1
𝑎
) −
𝑉𝑖𝑛1
𝑎
] − [𝑉𝑖𝑛1 (1 +
1
𝑎
) −
𝑉𝑖𝑛2
𝑎
]}
𝑉𝑜 =
𝑅1
𝑅4
(1 +
2
𝑎
) (𝑉𝑖𝑛2 − 𝑉𝑖𝑛1)
𝑉𝑜 =
10 k𝛺
1 k𝛺
(1 +
2
𝑎
) (−1 sen(𝜔2𝑡) V + 0.5 sen(𝜔1𝑡) V)
𝑉𝑜 = 10 × (1 +
2
𝑎
) (−1 sen(𝜔2𝑡) + 0.5 sen(𝜔1𝑡) ) V
𝑉− = 𝑉+
𝑉𝑑 → 0 V
𝐼+ = 𝐼− = 0 A
𝐿𝑒𝑦 𝑑𝑒 𝑛𝑜𝑑𝑜𝑠 𝑑𝑒 𝐾𝑖𝑟𝑐ℎℎ𝑜𝑓𝑓
𝑽− :
0 V − 𝑉−
𝑅𝐴
+
𝑉𝑜 − 𝑉−
𝑅𝐹
= 0 V 𝑬𝒄. 𝟏
𝑽+ :
𝑉1 − 𝑉+
𝑅
+
𝑉2 − 𝑉+
10𝑅
= 0 V 𝑬𝒄. 𝟐
𝑉− = 𝑉+
𝐸𝑐. 1:
−𝑉−
𝑅𝐴
+
𝑉𝑜 − 𝑉−
𝑅𝐹
= 0 V
𝑉− (
1
𝑅𝐴
+
1
𝑅𝐹
) =
𝑉𝑜
𝑅𝐹
𝑉− = 𝑉𝑜 (
𝑅𝐴
𝑅𝐴 + 𝑅𝐹
) 𝐸𝑐. 3
𝐸𝑐. 2:
𝑉1 − 𝑉+
𝑅
+
𝑉2 − 𝑉+
10𝑅
= 0 V
𝑉1
𝑅
−
𝑉2
10𝑅
= 𝑉+ (
1
𝑅
+
1
10𝑅
)
𝑉+ =
10𝑉1 − 𝑉2
11
𝐸𝑐. 4
𝐸𝑐. 3 𝑒𝑛 𝐸𝑐. 4
𝑉𝑜 (
𝑅𝐴
𝑅𝐴 + 𝑅𝐹
) =
10𝑉1 − 𝑉2
11
𝑽𝒐 =
𝟏
𝟏𝟏
(𝟏 +
𝑹𝑭
𝑹𝑨
) (𝟏𝟎𝑽𝟏 − 𝑽𝟐)
𝛽
𝑅𝐸
𝛼
𝛽 + 1
𝑉𝐶𝐶
𝐼𝐷 𝑉𝐺𝑆
𝑉𝐺𝑆 < 0
𝑍𝑖𝑛 𝑍𝑜𝑢𝑡
𝐴𝑣1 ≠ 𝐴𝑣2 𝐴𝑖1 ≠ 𝐴𝑖2 .
𝑉01 = 𝐴𝑣1𝑉𝑖1
𝑉02 = 𝐴𝑣2𝑉𝑖2
⇒ 𝑉𝑂𝑇 = 𝑉𝑖𝑛(𝐴𝑣1 ∗ 𝐴𝑣2 … ∗ 𝐴𝑣𝑛) [2.1]
AvT = ±𝐴𝑣1 ∗ Av2 ∗ … ∗ Avn [2.2]
𝐴𝑖𝑇 = ±𝐴𝑖1 ∗ 𝐴𝑖2 … ∗ 𝐴𝑖𝑛 [2.3]
APT =
Po
Pi
= AvtAit [2.4]
𝑉𝐵𝑄𝐷𝐶
𝑅𝐶
𝑋𝐶𝐵 𝑋𝐶𝐶
𝑋𝐸
𝑍𝑖𝑛
𝑍𝑜
𝐴𝑖
𝐴𝑣𝑇
𝒁𝒊𝒏
k3.3k4||k20
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1
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μA48
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83.10
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¿Cuánto vale el 𝑉𝑖𝑛𝑝 máximo que puede ingresar sin que se produzca
recortes?
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𝑍𝑖𝑛2
𝐴𝑣𝑇
𝑉𝑜
𝑍𝑜
𝑍𝑖𝑛1
𝒁𝒊𝒏𝟐
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iV
|𝐴𝑣| = 120
𝛽𝑚𝑖𝑛 = 100
𝑉𝑖𝑛 = 0.01 sen(𝜔𝑡) V
𝑓𝑚𝑖𝑛 = 1kHz
𝑅𝐿 = 4.7 kΩ
𝑍𝑖𝑛 ≥ 1.5 kΩ
𝑆𝑒 𝑑𝑒𝑏𝑒 𝑡𝑜𝑚𝑎𝑟 𝑒𝑛 𝑐𝑢𝑒𝑛𝑡𝑎 𝑞𝑢𝑒 𝐴𝑣1 > 𝐴𝑣2:
𝐴𝑣1 = 10 𝑦 𝐴𝑣2 = 12
kΩ35.2'
2
||'
kΩ7.4
2
2
L
L
LCL
LC
R
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' L
L
R
R
𝑉𝑅𝐶2 ≥
𝑅𝐶2
𝑅𝐿
′ × �̂�𝑜 =
4.7 kΩ
2.35 kΩ
× 120 × 10 mV = 2.4 V × 1.2
1.2 = 𝐹𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝑠𝑒𝑔𝑢𝑟𝑖𝑑𝑎𝑑 (10% 𝑜 20%)
𝐸𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑠𝑒𝑔𝑢𝑟𝑎 𝑞𝑢𝑒 𝑙𝑎 𝑠𝑒ñ𝑎𝑙 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑑𝑎 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒 𝑎 𝑙𝑎 𝑠𝑎𝑙𝑖𝑑𝑎.
V88.22 RCV
2
2
2
2 mA612.0
kΩ7.4
V88.2
E
C
RC
C I
R
V
I
Ω48.42
mA612.0
mV26
2
2
E
T
e
I
V
r
V12 inE VV
V2: 2 EVsi
Ω97.3267
mA612.0
V2
2 ER
22
2
'
'
Ee
L
v
Rr
R
A
150
160
35.15348.42
12
kΩ35.2'
' 2
2
2 e
L
E r
Av
R
R
150' 2ER
kΩ0.3
kΩ3.3
97.311715097.3267''' 222
EEE RRR
KR E 3'' 2
kΩ15.3kΩ3Ω150''' 222 EEE RRR
V92.1kΩ15.3mA612.0222 EEE RIV
𝑉𝐶𝐸2 = �̂�𝑖𝑛2 + �̂�𝑜 + 2 V = (10 mV × 10) + (10 mV × 120) + 2 V = 3.3 V
V1.8V88.2V3.3V92.1222 RCCEECC VVVV
V10CCV
μA12.6
100
mA612.02
2
C
B
I
I
μA2.61μA12.61010 222 BII
μA32.67μA12.61110 212 BII
kΩ39
kΩ43
kΩ81.42
μA2.61
V7.0V92.1
22
2
22
I
VV
R BEE
B
kΩ3922 BR
kΩ100
kΩ110
kΩ625.109
μA32.67
V)7.0V92.1(V10)(
12
2
21
I
VVV
R BEECC
B
kΩ11021 BR
kΩ44.19Ω)150Ω48.42101)')(1( 222 EeinT RrZ
kΩ791.28kΩ110||kΩ39|| 22212 BBBB RRR
kΩ6.11kΩ79.28||kΩ44.19222 BBinTin RZZ
kΩ11
kΩ12
21
inC ZR
kΩ111 CR
KKKRZR CinL 64.511//6.11||' 121
𝑉𝑅𝐶1 ≥
𝑅𝐶1
𝑅𝐿1
′ × �̂�𝑜
V195.010mV10
kΩ64.5
kΩ11
1 RCV
V31 RCV
1
1
1
1 mA27.0
kΩ11
V3
E
C
RC
C I
R
V
I
3.95
mA27.0
mV26
1
1
E
T
e
I
V
r
V2: 1 EVAsumir
kΩ3.7
mA27.0
V2
1
1
1
E
E
E
I
V
R
11
1
1
'
'
Ee
L
v
Rr
R
A
430
470
7.4683.95
10
kΩ64.5
' 1ER
470' 1ER
kΩ8.6
kΩ2.8
kΩ83.6470kΩ3.7''' 111
EEE RRR
kΩ2.8'' 1 ER
μA7.2
100
mA27.01
1
C
B
I
I
μA27μA7.21010 121 BII
μA7.29μA27.01111 111 BII
kΩ100
μA27
V7.0V2
21
1
21
I
VV
R BEE
B
kΩ10021 BR
kΩ240
kΩ270
kΩ8.245
μA7.29
V7.2V10)(
111
11
I
VVV
R BEECC
B
kΩ27011 BR
kΩ09.57)4703.95(101)')(1( 111 EeinT RrZ
kΩ97.72kΩ270||kΩ100|| 11211 BBBB RRR
kΩ03.32kΩ97.72||kΩ09.57|| 11 inTBBin ZRZ
𝐶𝐶1 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿1
′ ≥
10
2𝜋 × 1 kHz × 5.64 Ω
≥ 0.28 µF
𝑪𝑪 = 𝟏 µF
𝐶𝐸1 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸1
′ )
≥
10
2𝜋 × 1 kHz × (95.3 Ω + 470 Ω)
= 2.82 µF
𝑪𝑬 = 𝟑. 𝟑 µF
𝐶𝐵1 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛1
≥
10
2𝜋 × 1 kHz × 32.03 kΩ
= 0.05 µF
𝑪𝑩 = 𝟏 µF
𝐶𝐶2 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿2
′ ≥
10
2𝜋 × 1 kHz × 2.35 kΩ
≥ 0.68 µF
𝑪𝑪 = 𝟏 µF
𝐶𝐸2 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × (𝑟𝑒 + 𝑅𝐸2
′ )
≥
10
2𝜋 × 1 kHz × (42.48 Ω + 150 Ω)
= 8.27 µF
𝑪𝑬 = 𝟏𝟎 µF
𝐶𝐵2 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛2
≥
10
2𝜋 × 1 kHz × 11.6 kΩ
= 0.14 µF
𝑪𝑩 = 𝟏 µF
|𝐴𝑣1| =
5.64 kΩ
95.3 Ω + 470 Ω
= 9.98
|𝐴𝑣2| =
2.35 kΩ
42.48 Ω + 150 Ω
= 12.21
|𝑨𝒗𝑻| = |𝑨𝒗𝟏| ∙ |𝑨𝒗𝟐| = 𝟗. 𝟗𝟖 × 𝟏𝟐. 𝟐𝟏 = 𝟏𝟐𝟏. 𝟖
kΩ50.1kΩ03.32 inZ
Vi =10 mV
Av = 200
RL = 4.7 kΩ
Zint >100 kΩ
gm = 2.2 mS
IDSS = 4 mA
rd = 40 kΩ
Vp = -4 V
𝐼𝐶2 ≈ 𝐼𝐸2 ≈ 𝐼𝐶1 ≈ 𝐼𝐸1 𝑠𝑖 𝛽1 = 𝛽2 = 𝛽 ⟹ 𝐼𝐵1 = 𝐼𝐵2
𝑉𝐵1 = 𝑉𝑅𝐵3 = 𝑉𝐶𝐶
𝑅𝐵3
(𝑅𝐵3 + 𝑅𝐵2 + 𝑅𝐵1)
[2.5]
𝑉𝐵2 = 𝑉𝐶𝐶
(𝑅𝐵2 + 𝑅𝐵3)
(𝑅𝐵3 + 𝑅𝐵2 + 𝑅𝐵1)
[2.6]
𝑉𝐶2 = 𝑉𝐶𝐶 − 𝐼𝐶2 ∙ 𝑅𝐶 [2.7]
𝑉𝐶1=𝑉𝐸2 = 𝑉𝐵2 − 𝑉𝐵𝐸 [2.8]
𝐼𝐵1 =
𝑉𝐶𝐶
𝑅𝐵3
(𝑅𝐵3 + 𝑅𝐵2 + 𝑅𝐵1)
− 𝑉𝐵𝐸
(𝛽 + 1)𝑅𝐸
𝐼𝐵1 =
𝑉𝐸1
(𝛽+1)𝑅𝐸
[2.9]
𝐴𝑉𝑇 = 𝐴𝑣𝐸𝐶 ∙ 𝐴𝑣𝐵𝐶 [2.10]
𝐴𝑣𝐸𝐶 = −
𝑍𝑖𝑛2
𝑟𝑒1
[2.11]
𝐴𝑣𝐵𝐶 =
𝑅𝐿′
𝑍𝑖𝑛2
[2.12]
𝑍𝑖𝑛2 = 𝑟𝑒2
𝐴𝑣𝐸𝐶 = −
𝑟𝑒2
𝑟𝑒1
≈ 1 𝐵𝑎𝑗𝑎 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎𝑙 𝑒𝑓𝑒𝑐𝑡𝑜 𝑀𝑖𝑙𝑒𝑟 [2.13]
𝐴𝑣𝐵𝐶 =
𝑅𝐿′
𝑟𝑒2
[2.14]
𝐴𝑣𝑡 = (−
𝑟𝑒2
𝑟𝑒1
) (
𝑅𝐿′
𝑟𝑒2
) = −
𝑅𝐿′
𝑟𝑒1
[2.15]
𝑍𝑖𝑛 = 𝑍𝑖𝑛1 = 𝑅𝐵3 ∥ 𝑅𝐵2 ∥ 𝑍𝑖𝑛𝑇1 [2.16]
𝑍𝑖𝑛𝑇1 = (𝛽1 + 1)(𝑟𝑒1) [2.17]
𝑍𝑖𝑛2 = 𝑟𝑒2 +
𝑅𝐵1 ∥ 𝑅𝐵2
(𝛽2 + 1)
[2.18]
𝐴𝑣𝐸𝐶 =
𝑟𝑒2 +
𝑅𝐵1 ∥ 𝑅𝐵2
(𝛽2 + 1)
𝑟𝑒1
[2.19]
𝐴𝑣𝐵𝐶 =
𝑅𝐿′
𝑟𝑒2 +
𝑅𝐵1 ∥ 𝑅𝐵2
(𝛽2 + 1)
[2.20]
𝐴𝑣𝑇 = (−
𝑟𝑒2 +
𝑅𝐵1 ∥ 𝑅𝐵2
(𝛽2 + 1)
𝑟𝑒1
) (
𝑅𝐿′
𝑟𝑒2 +
𝑅𝐵1 ∥ 𝑅𝐵2
(𝛽2 + 1)
) = −
𝑅𝐿′
𝑟𝑒1
[2.21]
𝑍𝑖𝑛 = 𝑍𝑖𝑛1 = 𝑅𝐵3 ∥ 𝑅𝐵2 ∥ 𝑅𝐵1 ∥ 𝑍𝑖𝑛𝑇1 [2.22]
𝑍𝑖𝑛𝑇1 = (𝛽1 + 1)(𝑟𝑒1) [2.23]
RB2 .
𝐴𝑣𝑇 = 𝐴𝑣𝑆𝐶 ∙ 𝐴𝑣𝐵𝐶 [2.24]
𝐴𝑣𝑆𝐶 = −𝑔𝑚 ∗ 𝑟𝑒 [2.25]
𝐴𝑣𝐵𝐶 =
𝑅𝐿′
𝑟𝑒
[2.26]
𝐴𝑣𝑇 = −𝑔𝑚 ∙ 𝑟𝑒 ∙
𝑅𝐿′
𝑟𝑒
[2.27]
𝐴𝑣𝑇 = −𝑔𝑚 ∙ 𝑅𝐿′ [2.28]
max
max
in
o
Tv
V
V
A
LCo RIV 'max
kΩ65.1kΩ3.3||kΩ3.3||' LCL RRR
SDEC IIII
2
1
P
GS
DSSD
V
V
II
0GV
SDSGS RIVV
2
V4
kΩ1
1mA10
D
D
I
I
16
kΩ1
2
1
1mA10
22
DD
D
IKI
I
2
6255m10 DDD III
mA14.21 DI
V14.2kΩΩ1mA)(14.2(11 SDGS RIV
Se escoge esta respuesta por: |𝑉𝐺𝑆| < |𝑉𝑝|
mA45.71 DI
V45.72 GSV
V53.3kΩ65.1mA14.2max oV
𝐴𝑣 = −𝑔𝑚 × 𝑅′
𝐿
P
GS
P
DSS
V
V
V
I
gm 1
2
mS32.2
V4
V14.2
1
V4
mA102
mg
83.3kΩ65.1mS32.2
TvA
V92.0
83.3
V53.3max
max
vT
o
in
A
V
V
𝑉𝑜𝑝 +:
op
L
C
RC V
R
R
V
'
𝑉𝑜𝑝 −:
min222 CEipopCE VVVV
min111 CEipopCE VVVV
V11 EV
in
S
Zf
C
2
10
ref
CE
2
10
12
10
BB
B
Rf
C
211 || BBBB RRR
L
C
Rf
C
'2
10
𝑉0 = −2.5 sen(𝜔𝑡) V
𝛽𝑚𝑖𝑛 = 100
𝑉𝑖𝑛 = 0.02 sen(𝜔𝑡) V
𝑓𝑚𝑖𝑛 = 1kHz
𝑍𝑖𝑛 ≥ 2 kΩ
𝑅𝐿 = 2.2 kΩ
¿Es posible diseñar con estos datos?:
1231
|||| inTBBinin ZRRZZ
23 || BBBB RRR
11|
TinZinTRBBin ZZ
kΩ21
111
eTinin rZZ
8.19
101
kΩ2
1
kΩ2
min1
er
121
2
21
''
||||||
e
L
e
L
e
e
vvTv
r
R
r
R
r
r
AAA
1
'
||
e
L
Tv
r
R
A →
||
|'
| max
max1
Tv
L
e
A
R
r
𝐴𝑣𝑇 =
�̂�𝑜
�̂�𝑖𝑛
=
2.5 V
20 mV
= 125
6.17
125
kΩ2.2
||max1
Tv
L
e
A
R
r
max11min1 eee rrr
||1
1
Tv
L
e
in
A
R
r
Z
6.178.19
1e
r
No es posible realizar este diseño con los datos planteados.
Diseñar un amplificador cascode que cumpla con los siguientes requisitos:
|𝐴𝑣| = 50
𝛽𝑚𝑖𝑛 = 120
𝑉𝑖𝑛 = 0.01 sen(𝜔𝑡) V
𝑓𝑚𝑖𝑛 = 1 kHz
𝑅𝐿 = 2.2 kΩ
𝑍𝑖𝑛 ≥ 2 kΩ
¿Es posible diseñar con estos datos?
1231
||||
TinBBinin ZRRZZ
2323 || BBBB RRR
1123|
TinZinTRBBin ZZ
kΩ21
111
eTinin rZZ
52.16
121
kΩ2
1
kΩ2
min1
er
121
2
21
''
||||||
e
L
e
L
e
e
vvTv
r
R
r
R
r
r
AAA
1
'
||
e
L
Tv
r
R
A
||
|'
| max
max1
Tv
L
e
A
R
r
44
50
kΩ2.2
||max1
Tv
L
e
A
R
r
max11min1 eee rrr
||1
1
Tv
L
e
in
A
R
r
Z
4452.16
1e
r
Si es posible diseñar pero con un rango muy bajo.
𝑆𝑒 𝑒𝑠𝑐𝑜𝑔𝑒 𝑢𝑛𝑎 𝑟𝑒1 = 30𝛺
15005030
1 VeL ArR
kΩ7.4
kΩ1.5
kΩ77.4
kΩ5.1kΩ2.2
kΩ5.1kΩ2.2
'
'
LL
LL
C
RR
RR
R
kΩ7.4CR
kΩ5.1||' LCL RRR
𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅′𝐿
× �̂�𝑜
][76.32.1][13.35020
kΩ5.1
kΩ7.4
VVmVVRC
V4: RCVAsumir
E
C
RC
C I
R
V
I mA85.0
kΩ7.4
V4
55.30
mA85.0
mV26
21
C
T
ee
I
V
rr
V11 EV
𝑆𝑒𝑎 𝑉𝐸 = 2𝑉
kΩ2.2
kΩ4.2
kΩ35.2
mA85.0
V2
E
E
E
I
V
R
kΩ4.2ER
min222
ˆˆ
CEinoCE VVVV min211
ˆˆ
CEooCE VVVV
V2mV20mV20502 CEV
V2mV20mV201 CEV
V98.22 CEV
V04.21 CEV
ECECERCCC VVVVV 12
V11V2V2.04V2.98V4 CCV
Se estandariza en: VVCC 12
V7.2V7.01 EB VV
μA7
121
mA85.0
1
1
C
B
I
I
μA4.8412 11 BII
μA36.7711 12 BII
μA33.7010 13 BII
1
2
1
I
VVcc
R B
B
2
12
2
I
VV
R BB
B
3
1
3
I
V
R B
B
V74.4
V2V04.2V7.0
2
2
1122
B
BECEBEB
V
V
VVVV
μA4.84
V74.4V12
1
BR
μA36.77
V7.2V74.4
2
BR
μA33.70
V7.2
3 BR
kΩ82
kΩ91
kΩ861
BR
kΩ24
kΩ27
kΩ36.262
BR
kΩ36
kΩ39
kΩ383
BR
Estandarizando:
kΩ821 BR
kΩ272 BR
kΩ393 BR
kΩ7.355.30121))(1(1 einT rZ
kΩ7.3||kΩ95.15kΩ7.3||kΩ27||kΩ39|||| 123 inTBBin ZRRZ
kΩ3inZ
1.49
55.30
kΩ5.1'
1
e
L
v
r
R
A
μF53.0
kΩ3kHz12
10
2
10
min
in
S
Zf
C
μF1SC
μF52
55.30kHz12
10
2
10
min
e
E
rf
C
μF100EC
kΩ31.20|| 211 BBBB RRR
μF078.0
kΩ31.20kHz12
10
2
10
1min
BB
B
Rf
C
μF1.0BC
μF96.0
kΩ5.1kHz12
10
'2
10
min
L
C
Rf
C
μF1CC
𝐴𝑐𝑜𝑝𝑙𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑑𝑖𝑟𝑒𝑐𝑡𝑜 {
𝑄1 = 𝐶𝐶
𝑄2 = 𝐶
𝐼𝐵 = 𝐼𝐵1 [2.29]
𝐼𝐸 = 𝐼𝐸2 [2.30]
𝐼𝐸1 = 𝐼𝐵2 [2.31]
𝐼𝐶 = 𝐼𝐶1 + 𝐼𝐶2 = 𝛽1𝐼𝐵1 + 𝛽2𝐼𝐵2 = 𝛽1𝐼𝐵1 + 𝛽2𝐼𝐸1
𝐼𝐶 = 𝛽1𝐼𝐵1 + 𝛽2(1 + 𝛽1)𝐼𝐵1
𝐼𝐶 = 𝐼𝐵[𝛽1 + 𝛽2(1+𝛽1)] [2.32]
𝛽𝐷 =
𝐼𝐶
𝐼𝐵
= [𝛽1 + 𝛽2(1 + 𝛽1)] [2.33]
Si 𝛽1, 𝛽2>>1 𝛽𝐷 ≈ 𝛽1𝛽2 + 𝛽1
Si 𝛽1, 𝛽2>> 𝛽1 𝛽𝐷 ≈ 𝛽1𝛽2
Si 𝛽1 = 𝛽2 = 𝛽
𝛽𝐷 = 𝛽2 [2.34]
𝑟𝑒𝐷 =
𝑟𝑒1
(𝛽2+1)
+ 𝑟𝑒2 [2.35]
𝑟𝑒1 =
𝑉𝑇
𝐼𝐸1
𝑟𝑒2 =
𝑉𝑇
𝐼𝐸2
𝐼𝐸2 = (𝛽2 + 1)𝐼𝐸1 [2.36]
𝑟𝑒2 =
𝑉𝑇
(𝛽2 + 1)𝐼𝐸1
=
𝑟𝑒1
(𝛽2 + 1)
𝑟𝑒𝐷 = 𝑟𝑒2 + 𝑟𝑒2
𝑟𝑒𝐷 = 2𝑟𝑒2 [2.37]
Cálculo de DC y reDC .
𝐼𝐵 = 𝐼𝐵1 [2.38]
𝐼𝑐1 = 𝐼𝐵2 [2.39]
𝐼𝐶 = 𝐼𝐸2 = (1 + 𝛽2)𝐼𝐵2 = (1 + 𝛽2)𝐼𝐶1
𝐼𝐶 = (1 + 𝛽2)𝛽1𝐼𝐵1
𝐼𝐶 = (1 + 𝛽2)𝛽1𝐼𝐵 [2.40]
𝛽𝐷𝐶 ≡
𝐼𝐶
𝐼𝐵
= 𝛽1(1 + 𝛽2) [2.41]
Si 1, 2 >> 1 DC 12
Si 1 = 2 =
DC = 2 [2.42]
𝐼𝐸 = 𝐼𝐸1 + 𝐼𝐶2
𝐼𝐸 = 𝐼𝐸1 + 𝛽2𝐼𝐵2
𝐼𝐸 = 𝐼𝐸1 + 𝛽2𝐼𝐶1
𝐼𝐸 = 𝐼𝐸1 + 𝛽2𝛼1𝐼𝐸1
𝐼𝐸 = 𝐼𝐸1(1 + 𝛽2𝛼1)
𝐼𝐸
𝐼𝐸1
= (1 + 𝛽2𝛼1) [2.43]
Nota: La última fórmula indica que cualquier impedancia que hay en emisor
de Q1 puede ser transferida a emisor de Q2 dividiendo por (1 + 𝛽2𝛼1) .
𝑟𝑒𝐷𝐶 =
𝑟𝑒1+𝑅
1+𝛽2𝛼1
[2.44]
Si 11 y 2 1
𝑟𝑒𝐷𝐶 =
𝑟𝑒1+𝑅
𝛽2
[2.45]
Si 1 = 2 =
𝐴𝑖𝐷 = 𝛽2 = 𝛽𝐷 [2.46]
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒𝐷+𝑅𝐸
[2.47]
𝑍𝑖𝑛𝑇𝐷 = 𝛽𝐷(𝑟𝑒𝐷 + 𝑅𝐸); 𝐴𝑙𝑡𝑎 [2.48]
𝑍𝑖𝑛 = 𝑅𝐵𝐵||𝑍𝑖𝑛𝑇𝐷 ≈ 𝑅𝐵𝐵 [2.49]
𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅𝐿
′ ∙ 𝑉𝑜 [2.50]
𝑉𝐵𝐸 = 1.4𝑉 [2.51]
𝐼𝑅 ≪ 𝐼𝐵2 [2.52]
𝑅 =
𝑉𝐸+𝑉𝐵𝐸
𝐼𝐸1
10
[2.53]
Si 1 = 2 =
DAi 2
[2.54]
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒𝐷+𝑅𝐿′
[2.55]
𝑍𝑖𝑛𝑇𝐷 = 𝛽𝐷(𝑟𝑒𝐷 + 𝑅𝐿
′ ); 𝑎𝑙𝑡𝑎 [2.56]
𝑍𝑖𝑛 = 𝑅𝐵𝐵||𝑍𝑖𝑛𝑇𝐷 ≈ 𝑅𝐵𝐵 [2.57]
𝑋𝐶𝐵 ≪ 𝑍𝑖𝑛
𝑋𝐶𝐸 ≪ 𝑅𝐿
′
𝑉0 = 4 sen(𝜔𝑡) V
𝛽𝑚𝑖𝑛 = 100
𝑓𝑚𝑖𝑛 = 1 kHz
𝑍𝑖𝑛 ≥ 30 kΩ
𝑅𝐿 = 270 𝛺
𝐴𝑣 = 1 𝑐𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑐𝑖ó𝑛 𝐶– 𝐶
𝑆𝑖 𝛽1 = 𝛽2 = 𝛽 = 100
10000D
LeDDinTD RrZ '
BBinTDBBin RZRZ ||
𝑆𝑒𝑎 𝑹𝑬 = 𝑹𝑳 = 𝟐𝟕𝟎 𝜴
𝑅𝐿
′ = 135 𝛺
𝑉𝐸 ≥
𝑅𝐸
𝑅′
𝐿
× �̂�𝑜
[V]6.92.1[V]8
135
270
4
VVE
V10: EVAsumir
V4.112 BEEB VVV
2mA37
270
V10
ECD
E
E
ED II
R
V
I
(Se necesita un QD que permita esta corriente)
μA7.3
10000
mA37
D
CD
BD
I
I
kΩ300
kΩ330
kΩ308
μA7.310
V4.11
10
2
BD
B
B
I
V
R
KRB 3302
VVV ECC 202
kΩ200
kΩ220
kΩ211
μA7.311
V4.11V20
11
1
BD
BCC
B
I
VV
R
KRB 2201
kΩ132|| 21 BBBB RRR
4.1
mA37
mV2622
ED
T
eD
I
V
r
MΩ3.11354.110000' LeDDTD RrZin
kΩ353.120kΩ132||MΩ3.1 Zin
10
2
22
B
EBE
I
VV
R
kΩ27
kΩ30
kΩ9.28
10
μA7.3
7.010
R
KR 27
𝐶𝐸 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑅𝐿
′ ≥
10
2𝜋 × 1 kHz × 135 Ω
= 11.79 µF
𝑪𝑬 = 𝟐𝟐 µF
𝐶𝐵 ≥
10
2𝜋 × 𝑓𝑚𝑖𝑛 × 𝑍𝑖𝑛
≥
10
2𝜋 × 1 kHz × 120.353 kΩ
= 0.013 µF
𝑪𝑩 = 𝟏 µF
𝑍𝑜 = 𝑅𝐿
′ = 𝑅𝐸||𝑅𝐿 [2.58]
𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒𝐷𝐶+𝑅𝐿′
= 𝐴𝑣 =
𝑅𝐿
′
𝑟𝑒1+𝑅
𝛽2
+𝑅𝐿′
[2.59]
𝑍𝑖𝑛𝑇𝐷𝐶 = 𝛽𝐷𝐶(𝑟𝑒𝐷𝐶 + 𝑅𝐿
′ ) [2.60]
𝑍𝑖𝑛𝑇𝐷𝐶 = 𝛽1𝛽2 (
𝑟𝑒1 + 𝑅
𝛽2
+ 𝑅𝐿
′ )
𝑍𝑖𝑛𝑇𝐷𝐶 = 𝛽1(𝑟𝑒1 + 𝑅) + 𝛽1𝛽2𝑅𝐿
′ [2.61]
𝑅𝐵𝐵 = 𝑅1||𝑅2
𝑍𝑖𝑛 = 𝑅𝐵𝐵||𝑍𝑖𝑛𝑇𝐷𝑐 [2.62]
𝑅𝐵𝐵 = 𝑅𝐵1||𝑅𝐵2
𝐴𝑣 = −
𝑅𝐿
′
𝑟𝑒+𝑅𝐸
′ ||𝑅𝐵𝐵
[2.63]
𝑍𝑖𝑛 = 𝑍𝑖𝑛𝑇 = (𝛽 + 1)(𝑟𝑒 + 𝑅𝐸
′ ||𝑅𝐵𝐵) [2.64]
𝑍𝑜 = 𝑅𝐿
′ = 𝑅𝐶||𝑅𝐿 [2.65]
(𝑟𝑒 + 𝑅𝐸
′ ||𝑅𝐵𝐵)|𝑚𝑎𝑥 =
𝑅𝐿
′ |𝑚𝑎𝑥
|𝐴𝑣|
=
𝑅𝐿
|𝐴𝑣|
𝑦
(𝑟𝑒 + 𝑅𝐸
′ ||𝑅𝐵𝐵)|𝑚𝑖𝑛 =
𝑍𝑖𝑛
𝛽 + 1
1.
𝑍𝑖𝑛
𝛽+1
< (𝑟𝑒 + 𝑅𝐸
′ ||𝑅𝐵𝐵) <
𝑅𝐿
|𝐴𝑣|
2. 𝑉𝑅𝐶 ≥
𝑅𝐶
𝑅𝐿
′ ∙ 𝑉𝑜
3. 𝑉𝐶𝐸 = 𝑉𝑜𝑝 + 𝑉𝑖𝑝 + 𝑉𝐶𝐸𝑚𝑖𝑛
4. 𝑉𝐸1 = 1𝑉
5. 𝑅3 =
𝑉𝐵−(𝑉𝐵𝐸−𝑉𝐸)
𝐼𝐵
≈
0.1𝑉
𝐼𝐵
6. 𝐶1 ≥
10
2𝜋∙𝑓𝑚𝑖𝑛 ∙𝑅𝐵𝐵
7. 𝐶𝐵 ≥
10
2𝜋∙𝑓𝑚𝑖𝑛 ∙𝑍𝑖𝑛
8. 𝐶𝐸 ≥
10
2𝜋∙𝑓𝑚𝑖𝑛 (𝑟𝑒+𝑅𝐸
′ ||𝑅𝐵𝐵)
9. 𝐶𝐶 ≥
10
2𝜋∙𝑓𝑚𝑖𝑛 ∙𝑅𝐿
′
|𝐴𝑣| = 25
𝛽𝑚𝑖𝑛 = 70
𝑉𝑜 = 2 sen(𝜔𝑡) V
𝑓𝑚𝑖𝑛 = 1 kHz
𝑅𝐿 = 1.5 kΩ
𝑍𝑖𝑛 ≥ 2.2 kΩ
v
L
BBEe
in
A
R
RRr
Z '
||
1
)')(1( ERreZin
25
kΩ5.1
||
71
kΩ2.2 ' BBEe RRr
60||98.30 BBEe RRr
Sea 45|| BBEe RRr
11254525'LR
kΩ9.3
kΩ7.4
kΩ5.4
kΩ125.1kΩ5.1
kΩ125.1kΩ5.1
'
'
LL
LL
C
RR
RR
R
kΩ9.3CR
kΩ083.1||' LCL RRR
V9.71.1V2.7V2
kΩ083.1
kΩ9.3ˆ
'
o
L
C
RC V
R
R
V
V8: RCVAsumir
E
C
RC
C I
R
V
I mA05.2
kΩ9.3
V8
A
mAI
I C
B
89.28
70
05.2
675.12
mA05.2
mV26
C
T
e
I
V
r
𝑉𝐶𝐸 = �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 = 2 V + 80 mV + 2 V
V08.4CEV
𝑆𝑒𝑎 𝑉𝐸 = 2 V
975
mA05.2
V2
E
E
E
I
V
R
V7.2V7.0V2 BEEB VVV
kΩ1.9
kΩ10
kΩ34.9
10
V7.2
2
B
B
I
R
kΩ102 BR
V08.14V8V08.4V2 RCCEECC VVVV
V15CCVkΩ36
kΩ39
kΩ7.38
μA89.2811
V7.2V15
11
1
B
BCC
B
I
VV
R
kΩ391 BR
kΩ96.7|| 21 BBBB RRR
BBE
v
RR
A
||'675.12
kΩ083.1
25
kΩ30
kΩ33
76.30'
ER
33'ER
910
kΩ0.1
94233975'' ER
kΩ1'' ER
kΩ3.3
kΩ6.3
kΩ42.3
μA89.28
V1.0V1.0
3
BI
R
kΩ3.33 R
kΩ23.386.32675.1271)||')(1( BBEein RRrZ
78.23
86.32675.12
kΩ083.1
||'
'
BBEe
L
v
RRr
R
A
F1099.1
kΩ96.7kHz12
10
2
10 7
min
1
BBRf
C
μF22.01 C
μF49.0
kΩ23.3kHz12
10
2
10
min
in
B
Zf
C
μF1BC
μF84.34
33675.12kHz12
10
)||'(2
10
min
BBEe
E
RRrf
C
μF47EC
μF46.1
1083kHz12
10
'2
10
min
L
C
Rf
C
μF2.2CC
Q1 Q2
o
o
121
2
1
0
ooiinin
in
iin
VVVVV
V
VV
121
2
1
22
0
ooiinin
in
iin
VVVVV
V
VV
0021
2
1
oinin
iin
iin
VVV
VV
VV
En general:
𝑉𝑜 = 𝐴𝑚𝑐 ∙ 𝑉𝑖𝑛 𝑚𝑐 + 𝐴𝑚𝑑 ∙ 𝑉𝑖𝑛 𝑚𝑑 [2.66]
𝑉𝑑 = 𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 [2.67]
𝑉𝑚𝑐 =
𝑉𝑖𝑛1−𝑉𝑖𝑛2
2
[2.68]
𝑄1 = 𝑄2
𝐼𝐶 = 𝐼𝐶1 = 𝐼𝐶2 ≈ 𝐼𝐸1 = 𝐼𝐸2 = 𝐼𝐸 [2.69]
𝐼𝑅𝐸 = 2𝐼𝐸 [2.70]
𝑉𝐵1 = 𝑉𝐵2 = 0 [2.71]
𝑉𝐸1 = 𝑉𝐸2 = −0.7𝑉 [2.72]
𝐼𝑅𝐸 =
−0.7𝑉−(−𝑉𝐸𝐸)
𝑅𝐸
=
𝑉𝐸𝐸−0.7𝑉
𝑅𝐸
𝑉𝐶1 = 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝐼𝐶 ∙ 𝑅𝐶 [2.73]
𝑅𝐶1 = 𝑅𝐶2 = 𝑅𝐶 [2.74]
𝑉𝐶𝐸1 = 𝑉𝐶𝐸2 = 𝑉𝐶 − (−0.7𝑉) [2.75]
𝐼𝐵1 = 𝐼𝐵2 = 𝐼𝐵 =
𝐼𝐶
𝛽
[2.76]
𝑉𝐶 = 𝑉𝐶𝐶 −
(𝑉𝐸𝐸 − 0.7𝑉)𝑅𝐶
2𝑅𝐸
Ganancia en modo diferencial.
𝐴𝑣𝑚𝑑 =
𝑉𝑜𝑚𝑑
𝑉𝑖𝑛𝑚𝑑
[2.77]
𝐴𝑣𝐸𝐶 = −
𝑅𝐶1
𝑟𝑒1 + 𝑅𝐸||𝑟𝑒2
𝑟𝑒1, 𝑟𝑒2 ≪ 𝑅𝐸
𝐴𝑣𝐸𝐶 = −
𝑅𝐶
𝑟𝑒1+𝑟𝑒2
[2.78]
𝑆𝑖 𝑄1 = 𝑄2
𝑟𝑒1 ≈ 𝑟𝑒2 ≈ 𝑟𝑒
𝐴𝑣𝐸𝐶 = −
𝑅𝐶
2𝑟𝑒
[2.79]
𝐴𝑣𝐶𝐶 =
𝑅𝐿′
𝑟𝑒1 + 𝑅𝐿
′
𝐴𝑣𝐶𝐶 =
𝑅𝐸||𝑟𝑒2
𝑟𝑒 + 𝑅𝐸||𝑟𝑒2
𝐴𝑣𝐶𝐶 ≈
𝑟𝑒2
𝑟𝑒 + 𝑟𝑒2
=
𝑟𝑒2
2𝑟𝑒2
≈
𝑟𝑒
2𝑟𝑒
𝐴𝑣𝐶𝐶 =
1
2
[2.80]
𝐴𝑣𝐵𝐶 =
𝑅𝐶
𝑟𝑒 +
𝑅𝐵𝐵
(𝛽 + 1)
𝐴𝑣𝐵𝐶 =
𝑅𝐶2
𝑟𝑒||𝑅𝐸
≈
𝑅𝐶
𝑟𝑒
𝐴𝑣𝐵𝐶 =
𝑅𝐶
𝑟𝑒
[2.81]
𝐴𝑣𝑚𝑑 =
𝑉𝑜𝑚𝑑
𝑉𝑖𝑛𝑚𝑑
=
𝑉𝑜1
(𝑉𝑖𝑛1−𝑉𝑖𝑛2)
[2.82]
𝐴𝑣𝑚𝑑 = 𝐴𝑣𝐶𝐶 ∙ 𝐴𝑣𝐵𝑐 [2.83]
𝐴𝑣𝑚𝑑 = −
1
2
∙
𝑅𝐶
𝑟𝑒
𝐴𝑣𝑚𝑑 = −
𝑅𝐶
2𝑟𝑒
[2.84]
𝑉𝑜𝑑1 = −
𝑅𝐶
2𝑟𝑒
(𝑉𝑖𝑛1 − 𝑉𝑖𝑛2)
𝑉𝑜𝑑2 =
𝑅𝐶
2𝑟𝑒
(𝑉𝑖𝑛1 − 𝑉𝑖𝑛2) [2.85]
𝐴𝑣𝑚𝑐 =
𝑉𝑜𝑚𝑐
𝑉𝑖𝑛𝑚𝑐
[2.86]
𝑉𝑚𝑐 =
𝑉𝑖𝑛1+𝑉𝑖𝑛2
2
[2.87]
Si Vin1=Vin2 , la salida es igual a cero.
𝑉𝑚𝑐 =
𝑉𝑖𝑛1+𝑉𝑖𝑛2
2
=
𝑉𝑖+𝑉𝑖
2
= 𝑉𝑖 [2.88]
∆𝑖𝑅𝐸 = 2∆𝑖𝑒 = 2∆𝑖𝐶 [2.89]
∆𝑖𝐶 =
∆𝑖𝑅𝐸
2
=
∆𝑉𝑖𝑛
2𝑅𝐸
[2.90]
∆𝑉𝑜 = ∆𝑖𝑐 ∙ 𝑅𝐶 =
∆𝑉𝑖𝑛
2𝑅𝐸
∙ 𝑅𝐶 [2.91]
𝐴𝑚𝑐 =
∆𝑉𝑜
∆𝑉𝑖𝑛
= −
𝑅𝐶
2𝑅𝐸
𝐴𝑚𝑐 = −
𝑅𝐶
2𝑅𝐸
[2.92]
|𝐶𝑀𝑅𝑅| =
𝐴𝑚𝑐
𝐴𝑚𝑑
[2.93]
|𝐶𝑀𝑅𝑅|𝑖𝑑𝑒𝑎𝑙 =
𝐴𝑚𝑐
𝐴𝑚𝑑
= ∞ [2.94]
|CMRR| por ejemplo 100-120 el típico.
|CMRR| dB=20log|CMRR|
Se asume que Vin1 existe y Vin2 = 0
𝑉𝑖𝑛 𝑚𝑑 = 𝑉𝑖𝑛1 + 0 = 𝑉𝑖𝑛1
𝑉𝑜1 = 𝑉𝑖𝑛1 (
−𝑅𝐶
𝑟𝑒+𝑅𝐸||(𝑟𝑒+
𝑅𝐵2
𝛽+1
)
) [2.95]
𝑍𝑖𝑛2 = 𝑟𝑒2 +
𝑅𝐵2
𝛽+1
[2.96]
𝑍𝑖𝑛1 = 𝑅𝐵1 + [(𝛽 + 1) (𝑟𝑒1 + 𝑅𝐸|| (𝑟𝑒2 +
𝑅𝐵1
𝛽+1
))] [2.97]
Si:
i
E
C
i
E
C
mdmc
in
iin
V
r
R
V
R
R
V
VAVAV
V
VV
220
0
0
2
1 mcin mcin
i
E
C
mc
iin
iin
V
R
R
V
VAV
VV
VV
2
0
mcin 0
2
1
i
E
C
md
iin
iin
V
r
R
V
VAV
VV
VV
2
0
mdin 0
2
1
𝑍𝑖𝑛1 = 𝑅𝐵1 + [(𝛽 + 1) (𝑟𝑒1 + 𝑅𝐸|| (𝑟𝑒2 +
𝑅𝐵1
𝛽+1
))] [2.101]
𝐼𝐸 =
𝑉𝑅𝐸
𝑅𝐸
[2.102]
𝐼𝐸 =
𝑉𝑧−𝑉𝐵𝐸
𝑅𝐸
[2.103]
𝑉𝑅𝐸 = 𝑉𝑍 − 𝑉𝐵𝐸 [2.104]
𝐼𝐶 = 𝐼𝐸 = 𝐼𝐸1 + 𝐼𝐸2 = 2𝐼𝐸2 = 2𝐼𝐸1 [2.105]
RE .
𝐴𝑣𝑑 = 12
𝛽𝑚𝑖𝑛 = 100
𝑉𝑖𝑛𝑑 = 0.09 𝑠𝑖𝑛(𝜔𝑡) V
𝐶𝑀𝑅𝑅 > 65 dB
𝑍𝑖𝑛 ≥ 3.2 kΩ
a) Diseño con los emisores polarizados con una RE .
QQQ 21
EEECCC IIIIII 2121
ERE II 2
21 eee rrr
mV
IR
I
V
R
r
R
A CC
E
T
C
e
C
dv
262
2
2
mV
IR
A CC
dv
52
VmAmVAvIR dCC 624.0521252
VIRV CCRC 624.0
ind
od
vd
V
V
A
VmVVAV indvdod 08.19012
VVRC 08.1
No se puede diseñar. Se sugiere cambiar la ganancia agregando al circuito
RE1 y RE2 (RE1=RE2) como se muestra en la figura 2.28.
12 Ee
C
dv
Rr
R
A
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒 𝑹𝑪 = 𝟐. 𝟐𝑲𝜴 𝑦 𝑢𝑛 𝑉𝑅𝐶 > 1.08𝑉
VVRC 2
mA
K
V
R
V
I
C
RC
C 9.0
2.2
2
6.28
9.0
26
mA
mV
I
V
r
C
T
e
K
KK
r
A
R
R e
vd
C
E
62
68
8.626.28
122
2.2
2
1
621ER
VmVVVVVV CEinoCE 29008.1ˆˆ
min
VVCE 17.3
EEERE RmARIV 9.02
mcv
dv
A
A
CMRR E
C
mcv
R
R
A
2
65
212
C
E
R
R
CMRR
K
K
K
KR
R C
E
1.5
2.6
9.5
212
2.265
212
65
KRE 2.6
VKmAVRE 27.112.69.02
EBCERCCC VVVV
VVVVCC 7.017.32
VVCC 47.4
BEREREEE VVVV 1
VmVVVEE 7.036.5627.11
VVEE 12
Se selecciona CCEE VVV 12
eEEEein rRRRrZ 11 ||1
6.2862||2.6626.281100 KZin
KZin 17.18
Se asume: Datos del Diodo Zener
𝑉𝑍 = 5.1𝑉
𝐼𝑍 = 10𝑚𝐴
REBEQZ VVV 3
3BEQZRE VVV
VVVVRE 4.47.01.5
ZEERZ VVV
VVVVRZ 9.61.512
680
750
76.688
100
8.1
10
9.6
mA
mA
V
I
V
R
RZ
RZ
Z
680ZR
K
K
K
mA
V
I
V
R
RE
RE
E
4.2
7.2
42.2
8.1
4.4
KRE 4.2
𝐴𝑣𝑑 = 15
𝛽𝑚𝑖𝑛 = 100
𝑉𝑖𝑛𝑑 = 0.1 𝑠𝑖𝑛(𝜔𝑡) V
𝐶𝑀𝑅𝑅 > 36 dB
𝑍𝑖𝑛 ≥ 3.7 kΩ
QQQ 21
EEECCC IIIIII 2121
ERE II 2
21 eee rrr
mV
IR
I
V
R
r
R
A CC
E
t
C
e
C
dv
262
2
2
mV
IR
A CC
dv
52
VmAmVAvIR dCC 78.0521552
VIRV CCRC 78.0
𝐶𝑜𝑛𝑑𝑖𝑐𝑖ó𝑛 𝑑𝑒 𝑟𝑒𝑐𝑜𝑟𝑡𝑒: 𝑉𝑅𝐶 𝑉𝑜𝑝
ind
od
dv
V
V
A
VmVVAV inddvod 08.19012
VVRC 5.1
No se puede diseñar. Se sugiere cambiar la ganancia agregando al
circuito RE1 y RE2
RE 1 RE 2
12 Ee
C
dv
Rr
R
A
𝑆𝑒 𝑎𝑠𝑢𝑚𝑒 𝑢𝑛𝑎 𝑹𝑪 = 𝟐. 𝟕𝑲𝜴 𝑦 𝑢𝑛 𝑉𝑅𝐶 > 1.5𝑉
VVRC 2
mA
K
V
R
V
I
C
RC
C 74.0
7.2
2
1.35
74.0
26
mA
mV
I
V
r
C
RC
e
51
56
9.541.35
152
7.2
2
1
K
r
Av
R
R e
d
C
E
561ER
𝑉𝐶𝐸 = �̂�𝑜 + �̂�𝑖𝑛 + 𝑉𝐶𝐸𝑚𝑖𝑛 = 1.5𝑉 + 100𝑚𝑉 + 2𝑉
VVCE 6.3
EEERE RmARIV 74.02
CMRRCMRRdB 10log20
CMRR
dBCMRRdB
10log
20
36
20
096.63CMRR
mcv
dv
A
A
CMRR E
C
mcv
R
R
A
2
096.63
215
C
E
R
R
CMRR
K
K
K
KR
R C
E
6.5
2.6
67.5
215
7.2096.63
215
096.63
KRE 2.6
VKmAVRE 17.92.674.02
eEEEein rRRRrZ 11 ||1
1.3556||2.6561.351100 KZin
KZ
ni
26.18
BECERCCC VVVV
VVVVCC 7.06.32
VVCC 9.4
BEREREEE VVVV 1
VmAVVEE 7.074.05617.9
VVEE 91.9
Se selecciona CCEE VVV 10
Aunque no es estándar, para efectos teóricos está correcto.
100 : IDSS = 6 mA,
rd = 40 k Vp = –4 V,
5 V 32 V
Av = 100,
Vin = 0.1sen (ωt) V, Zin 10 k, RL = 2.2 k y la fMIN=20 kHz.
𝐺 = 𝑙𝑜𝑔10
𝑃2
𝑃1
[𝑏𝑒𝑙] [3.1]
𝐺𝑑𝐵 = 10𝑙𝑜𝑔10 (
𝑃2
𝑃1
) Para la relación entre potencias.
𝐺𝑑𝐵 = 20𝑙𝑜𝑔10 (
𝑉2
𝑉1
) Para la relación entre voltajes.
1mW
𝑍𝑜 = 600 Ω
𝑃 =
𝑉2
𝑍
⇒ 𝑉 = 775 mV
1mW
GdBm = 10log10 (
P
1 mW
)
𝑃𝑟 = 𝑃𝑡𝐺𝑡𝐺𝑟 (
𝜆
4𝜋𝑅
)
2
, [3.2]
donde:
R = distancia en metros desde el transmisor al receptor.
𝝀 = longitud de onda en metros.
(
𝝀
𝟒𝝅𝑹
)
𝟐
= pérdida que sufre la señal en el espacio libre.480 MHz
12 dB 50 Ω 100 km
80dB
𝜆 =
𝑐
𝑓
=
3 ∗ 108 (
m
s
)
480 MHz
= 0.625 m
(
𝑃𝑜
𝑃𝑖
)
𝑑𝐵
= (
𝑃𝑅𝑋
𝑃𝑇𝑋
)
𝑑𝐵
= 10 𝑙𝑜𝑔10 𝐺𝑡 + 10 𝑙𝑜𝑔10 𝑃𝑒𝑙 + 10 𝑙𝑜𝑔10 𝐺𝑟
(
𝑃𝑅𝑋
𝑃𝑇𝑋
)
𝑑𝐵
= 10 𝑙𝑜𝑔10 𝐺𝑡 + 10 𝑙𝑜𝑔10 𝐺𝑟 + 20 𝑙𝑜𝑔10 (
𝜆
4𝜋𝑅
)
(
𝑃𝑅𝑋
𝑃𝑇𝑋
)
𝑑𝐵
= 𝐺𝑡𝑑𝐵 + 𝐺𝑟𝑑𝐵 + 20 𝑙𝑜𝑔10 (
𝜆
4𝜋𝑅
)
(
𝑃𝑅𝑋
𝑃𝑇𝑋
)
𝑑𝐵
= 12𝑑𝐵 + 80𝑑𝐵 + 20 𝑙𝑜𝑔10 (
0.625 𝑚
4𝜋 ∗ 100 𝑘𝑚
)
(
𝑃𝑅𝑋
𝑃𝑇𝑋
)
𝑑𝐵
= 92𝑑𝐵 − 126𝑑𝐵 = −34𝑑𝐵
10 𝑙𝑜𝑔10 (
𝑃𝑅𝑋
𝑃𝑇𝑋
) = −34𝑑𝐵
𝑃𝑅𝑋
𝑃𝑇𝑋
= 0.398 ∗ 10−3
𝑃𝑅𝑋
𝑃𝑇𝑋
=
𝑉𝑅𝑋
2 ∗ 𝑅
𝑉𝑇𝑋
2 ∗ 𝑅
= (
𝑉𝑅𝑋
𝑉𝑇𝑋
)
2
= 0.398 ∗ 10−3
𝑉𝑅𝑋
𝑉𝑇𝑋
= √0.398 ∗ 10−3 = 19.9 ∗ 10−3
Se considera 𝑉𝑇𝑋 = 1𝑉
𝑉𝑅𝑋 = 1 𝑉 ∗ 19.9 ∗ 10
−3 = 19.9 𝑚𝑉
𝑓𝐿 y 𝑓𝐻 = frecuencias de corte de 1/2 potencia.
AB = ancho de banda = 𝑓𝐿 − 𝑓𝐻 [3.3]
𝑃𝑜 𝑚𝑒𝑑𝑖𝑎 = |
𝑉0
2
𝑅0
| = |
(𝐴𝑣 𝑚𝑒𝑑𝑖𝑎 × 𝑉𝑖)
2
𝑅𝑜
| [3.4]
𝑃𝑜(𝑓𝐿 ,𝑓𝐻) = |
(0.707𝐴𝑣 𝑚𝑒𝑑𝑖𝑎 × 𝑉𝑖)
2
𝑅𝑜
|
= 0.5 |
(𝐴𝑣 𝑚𝑒𝑑𝑖𝑎 × 𝑉𝑖)
2
𝑅𝑜
| = 0.5𝑃𝑜𝑚𝑒𝑑𝑖𝑎
𝑃𝑜(𝑓𝐿 ,𝑓𝐻)
=
1
2
𝑃𝑜𝑚𝑒𝑑𝑖𝑎 [3.5]
𝐴𝑣
𝐴𝑣 𝑚𝑒𝑑𝑖𝑎
(
𝐴𝑣
𝐴𝑣 𝑚𝑒𝑑𝑖𝑎
)
𝑑𝐵
= 20𝑙𝑜𝑔 |
𝐴𝑣
𝐴𝑣 𝑚𝑒𝑑𝑖𝑎
|
𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
=
𝑉2
𝑉1
𝐴𝑣 =
𝑅
𝑅 − 𝑗
1
𝜔𝐶
𝐴𝑣 =
𝑅
𝑅 − 𝑗
1
2𝜋𝑓𝐶
𝐴𝑣 =
𝑅
√𝑅2 + (
1
𝜔𝐶
)
2
∠𝜙
|𝐴𝑣| =
𝑅
√𝑅2+(
1
𝜔𝐶
)
2
[3.6]
R = XC
|𝐴𝑣| =
𝑅
𝑅√2
= 0.707 [3.7]
R = XC
|𝐴𝑣| = 20log10 (
1
√2
) = −3 dB [3.8]
𝑅 =
1
2𝜋𝑓𝐿𝐶
→ 𝑓𝐿 =
1
2𝜋𝑅𝐶
[3.9]
𝐴𝑣 =
𝑅
𝑅 − 𝑗
1
2𝜋𝑓𝐶
=
1
1 − 𝑗
1
2𝜋𝑅𝐶𝑓
=
1
1 − 𝑗
𝑓𝐿
𝑓
Av =
1
√1+(
fL
f
)
2
⏟
módulo
|tg−1 (
fL
f
)
⏟
fase
[3.10]
|𝐴𝑣|dB = 20log10
1
√1 + (
𝑓𝐿
𝑓
)
2
= −20log10 (1 + (
𝑓𝐿
𝑓
)
2
)
1
2⁄
|𝐴𝑣|dB = −10log10 [1 + (
𝑓𝐿
𝑓
)
2
] [3.11]
𝑆𝑖 𝑓 ≪ 𝑓𝐿 ⟶ (
𝑓𝐿
𝑓
)
2
≫ 1
|𝐴𝑣|dB = −20log10 [(
𝑓𝐿
𝑓
)] [3.12]
𝑓 ≪ 𝑓𝐿 ,
𝒂) 𝑆𝑖 𝑓 = 𝑓𝐿 ⟶
𝑓𝐿
𝑓
= 1
|𝐴𝑣|𝑑𝐵 = −20 log(1) = 0 dB
𝒃) 𝑆𝑖 𝑓 = 0.5𝑓𝐿 ⟶
𝑓𝐿
𝑓
= 2
|𝐴𝑣|𝑑𝐵 = −20 𝑙𝑜𝑔(2) ≈ −6 dB
𝒄) 𝑆𝑖 𝑓 = 0.25𝑓𝐿 ⟶
𝑓𝐿
𝑓
= 4
|𝐴𝑣|𝑑𝐵 = −20 𝑙𝑜𝑔(4) ≈ −12 dB
𝒅) 𝑆𝑖 𝑓 = 0.1𝑓𝐿 ⟶
𝑓𝐿
𝑓
= 10
|𝐴𝑣|𝑑𝐵 = −20 log(10) ≈ −20 dB
CB.
𝐴𝑣 =
𝑉2
𝑉1
=
𝑍𝑖𝑛
𝑍𝑖𝑛 + 𝑗
1
2𝜋𝑓𝐶𝐵
𝐴𝑣 =
𝑉2
𝑉1
=
1
1 + 𝑗
1
2𝜋𝑓𝐶𝐵𝑍𝑖𝑛
𝑆𝑖:
𝑓𝐿𝐶𝐵 =
1
2𝜋𝐶𝐵𝑍𝑖𝑛
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝐶𝐵 [3.13]
𝐴𝑣 =
𝑉2
𝑉1
=
1
1 + 𝑗
𝑓𝐿𝐶𝐵
𝑓
𝑍𝑖𝑛 =
1
2𝜋𝑓𝐶𝐵
𝑓𝐿𝐶𝐵 =
1
2𝜋𝐶𝐵𝑍𝑖𝑛
G :
𝐴𝑣 =
𝑉2
𝑉1
=
𝑍𝑖𝑛
(𝑍𝑖𝑛 + 𝑅𝐺) + 𝑗
1
2𝜋𝑓𝐿𝐶𝐵𝐶𝐵
𝑓𝐿𝐶𝐵 =
1
2𝜋𝐶𝐵(𝑍𝑖𝑛+𝑅𝐺)
[3.14]
RL
𝐴𝑣 =
𝑉2
𝑉1
=
𝑅𝐿
(𝑅𝐿 + 𝑅𝑐) + 𝑗
1
2𝜋𝑓𝐿𝐶𝐶𝐶𝐶
(𝑅𝐿 + 𝑅𝑐) =
1
2𝜋𝑓𝐿𝐶𝐶𝐶𝐶
𝑓𝐿𝐶𝐶 =
1
2𝜋𝐶𝐶(𝑅𝐿+𝑅𝑐)
𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝐶𝐶[3.15]
|𝐴𝑣| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸 ∥ 𝑋𝐶𝐸
𝑅𝐸 ∥ 𝑋𝐶𝐸 ≈ 𝑋𝐶𝐸 ⟶ 𝑅𝐸 ≫ 𝑋𝐶𝐸
|𝐴𝑣| =
𝑅𝐿′
𝑟𝑒 − 𝑗𝑋𝐶𝐸
=
𝑅𝐿′
𝑟𝑒 − 𝑗
1
2𝜋𝑓𝐿𝐶𝐸𝐶𝐸
𝑟𝑒 =
1
2𝜋𝑓𝐿𝐶𝐸𝐶𝐸
𝑓𝐿𝐶𝐸 =
1
2𝜋𝑟𝑒𝐶𝐸
𝐹𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝐶𝐸 [3.16]
RE RE’ RE” ,
|𝐴𝑣| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ + 𝑅𝐸
′′ ∥ 𝑋𝐶𝐸
𝑋𝐶𝐸 ∥ 𝑅𝐸
′′ ≈ 𝑋𝐶𝐸 ⟶ 𝑅𝐸
′′ ≫ 𝑋𝐶𝐸
|𝐴𝑣| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ − 𝑗𝑋𝐶𝐸
=
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ − 𝑗
1
2𝜋𝑓𝐿𝐶𝐸𝐶𝐸
𝑟𝑒 + 𝑅𝐸
′ =
1
2𝜋𝑓𝐿𝐶𝐸𝐶𝐸
𝑓𝐿𝐶𝐸 =
1
2𝜋𝐶𝐸(𝑟𝑒+𝑅𝐸
′)
𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑑𝑒𝑏𝑖𝑑𝑜 𝑎 𝐶𝐸 [3.17]
|𝐴𝑣| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ +
𝑅𝐸
′′ ∙ 𝑋𝐶𝐸
𝑅𝐸
′′ + 𝑋𝐶𝐸
=
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ +
𝑅𝐸
′′
1 +
𝑅𝐸
′′
𝑋𝐶𝐸
𝑓 ⟶ ∞ |𝐴𝑣 𝑚𝑎𝑥| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸 ′
𝑋𝐶𝐸 =
1
𝜔𝐶𝐸
= 0
𝑓 ⟶ 0 |𝐴𝑣 𝑚𝑖𝑛| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
′ + 𝑅𝐸
′′ 𝑋𝐶𝐸 =
1
𝜔𝐶𝐸
→ ∞
|𝐴𝑣 𝑚𝑖𝑛| =
𝑅𝐿′
𝑟𝑒 + 𝑅𝐸
𝑓𝐿𝐶𝐸
′ =
1
2𝜋𝐶𝐸(𝑟𝑒 + 𝑅𝐸)
[3.18]
𝑓𝐿𝐶𝐺 =
1
2𝜋𝐶𝐺(𝑅𝑔+𝑅𝐺)
[3.19]
𝑓𝐿𝐶𝐷 =
1
2𝜋𝐶𝐷(𝑅𝐷∥𝑟𝑑+𝑅𝐿)
[3.20]
𝑓𝐿𝐶𝐺 =
1
2𝜋𝐶𝑆𝑅𝑒𝑞
[3.21]
𝑅𝑒𝑞 = 𝑅𝑆 ∥
1
𝑔𝑚
𝐴𝑣 =
𝑉0
𝑉𝑖𝑛
=
−𝑗
1
2𝜋𝑓𝐶
𝑅 − 𝑗
1
2𝜋𝑓𝐶
|𝐴𝑣| =
𝑋𝑐
√𝑅2 + 𝑋𝐶
2
𝑆𝑖 𝑋𝑐 = 𝑅 ⇒ |𝐴𝑣| =
𝑋𝑐
𝑋𝑐√2
=
1
√2
[3.22]
𝑅 =
1
2𝜋𝑓𝐶
𝑓𝐻 =
1
2𝜋𝑅𝐶
[3.23]
|𝐴𝑣| =
1
1 + 𝑗𝑅2𝜋𝑓𝐶
=
1
1 +
𝑓
𝑓𝐻
𝐴𝑣 =
1
√1+(
𝑓
𝑓𝐻
)
2
⏟
𝑚ó𝑑𝑢𝑙𝑜
|𝑡𝑔−1 (
𝑓
𝑓𝐻
)
⏟
𝑓𝑎𝑠𝑒
[3.24]
Cbc, Cbe, Cce
Cwi Cwo
Cbc
𝐶𝑖𝑛 = 𝐶𝑤𝑖 + 𝐶𝑏𝑒 + 𝐶𝑀𝑖
𝐶𝑀𝑖 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 𝑒𝑛𝑡𝑟𝑎𝑑𝑎 = (1 − 𝐴𝑣)𝐶𝑓
𝐶𝑓 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑖𝑎 𝑑𝑒 𝑟𝑒𝑡𝑟𝑜𝑎𝑙𝑖𝑚𝑒𝑛𝑡𝑎𝑐𝑖ó𝑛
𝑑𝑒 𝑠𝑎𝑙𝑖𝑑𝑎 𝑎 𝑙𝑎 𝑒𝑛𝑡𝑟𝑎𝑑𝑎 = 𝐶𝑏𝑐
𝐶𝑖𝑛 = 𝐶𝑤𝑖 + 𝐶𝑏𝑒 + (1 − 𝐴𝑣)𝐶𝑏𝑐 [3.25]
Rthin = RB1 ∥ RB2 ∥ ZinT = Zin [3.26]
𝐴𝑣 =
𝑉𝑜
𝑉𝑡ℎ𝑖𝑛
=
−𝑗
1
𝜔𝐶𝑖𝑛
𝑅𝑡ℎ𝑖𝑛 − 𝑗
1
𝜔𝐶𝑖𝑛
⇒ 𝑅𝑡ℎ𝑖𝑛 =
1
𝜔𝐶𝑖𝑛
=
1
2𝜋𝑓𝐻𝑖𝐶𝑖𝑛
𝑓𝐻𝑖 =
1
2𝜋𝑅𝑡ℎ𝑖𝑛𝐶𝑖𝑛
[3.27]
Fig.3.20 Influencias de las Capacitancias de Salida.
𝐶𝑜=𝐶𝑤𝑜 + 𝐶𝑐𝑒 + 𝐶𝑀𝑜
𝐶𝑀𝑜 = 𝐶𝑎𝑝𝑎𝑐𝑖𝑑𝑎𝑑 𝑚𝑖𝑙𝑙𝑎𝑟 𝑑𝑒 𝑠𝑎𝑙𝑖𝑑𝑎 = (1 −
1
𝐴𝑣
)𝐶𝑓
𝐶𝑜 = 𝐶𝑤𝑜+𝐶𝑐𝑒 + (1 −
1
𝐴𝑣
)𝐶𝑏𝑐 [3.28]
𝐶𝑀𝑜 = (1 −
1
𝐴𝑣
)𝐶𝑏𝑐|
𝐴𝑣≫1
≈ 𝐶𝑏𝑐
𝑅𝑡ℎ𝑜 = 𝑅𝑐 ∥ 𝑅𝐿 ∥ 𝑍𝑂𝑇 = 𝑅𝐿
′ [3.29]
𝐴𝑣 =
𝑉𝑜
𝑉𝑖𝑛
=
𝑉𝑜
𝑉𝑡ℎ𝑜
=
−𝑗
1
𝜔𝐶𝑜
𝑅𝑡ℎ𝑜 − 𝑗
1
𝜔𝐶𝑜
𝑅𝑡ℎ𝑜 =
1
𝜔𝐶𝑜
𝑓𝐻𝑜 =
1
2𝜋𝑅𝑡ℎ𝑜𝐶𝑜
[3.30]
ℎ𝑓𝑒 =
ℎ𝑓𝑒 𝑚𝑒𝑑𝑖𝑎𝑠
1 + 𝑗
𝑓
𝑓𝛽
[3.31]
𝑓𝛽 = 𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑟𝑡𝑒 𝑑𝑒 𝑙𝑎 𝑟𝑒𝑠𝑝𝑢𝑒𝑠𝑡𝑎 𝑑𝑒 𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎 𝛽
ℎ𝑓𝑒 𝑚𝑒𝑑𝑖𝑎: ℎ𝑓𝑒 𝑒𝑛 𝑚𝑒𝑑𝑖𝑎𝑠 𝑓𝑟𝑒𝑐𝑢𝑒𝑛𝑐𝑖𝑎𝑠
ℎ𝑓𝑒 𝑛𝑜𝑚𝑖𝑛𝑎𝑙
𝑖𝑐
𝑖𝑏
= ℎ𝑓𝑒 = ℎ𝑓𝑒 𝑚𝑒𝑑𝑖𝑎𝑠
1
1 + 𝑗𝜔ℎ𝑖𝑒𝐶𝑒𝑞
𝐶𝑒𝑞 = 𝐶𝑏𝑒 + 𝐶𝑏𝑐
1 + 𝑗𝜔ℎ𝑖𝑒𝐶𝑒𝑞 = 0
𝑓𝛽 =
1
2𝜋ℎ𝑖𝑒𝐶𝑒𝑞
[3.32]
ℎ𝑖𝑒 = (𝛽 + 1)𝑟𝑒
ℎ𝑖𝑒 = 𝛽𝑟𝑒
𝑓𝛽 ≈
1
2𝜋𝛽𝑟𝑒𝐶𝑒𝑞
[3.33]
ℎ𝑓𝑒 = ℎ𝑓𝑒 𝑚𝑒𝑑𝑖𝑎𝑠
1
1 + 𝑗
𝑓
𝑓𝛽
|ℎ𝑓𝑒|𝑑𝐵 = 20𝑙𝑜𝑔|ℎ𝑓𝑒|
|ℎ𝑓𝑒|𝑑𝐵 = 20𝑙𝑜𝑔
(
ℎ𝑓𝑒 𝑚𝑒𝑑𝑖𝑎𝑠
√1 + (
𝑓
𝑓𝛽
)
2
)
[3.34]
f = ancho de banda
fT = frecuencia a la cual hfe = 1 no existe ganancia de corriente, puede
considerase como un límite de trabajo.
𝑆𝑖 𝑓 = 𝑓𝑇
ℎ𝑓𝑒 = 1 → 𝑓𝑇 = 𝛽𝑚𝑒𝑑𝑖𝑎𝑠𝑓𝛽
𝑓𝑇 =
1
2𝜋𝑟𝑒𝐶𝑒𝑞
[3.35]
𝑓𝛼 =
1
2𝜋𝐶𝑒𝑞𝐵𝐶
𝐶𝑒𝑞𝐵𝐶 = 𝐶𝑏𝑒𝑒
𝑓𝛽 =
1
2𝜋(𝛽 + 1)𝑟𝑒𝐶𝑒𝑞𝐸𝐶
𝐶𝑒𝑞𝐸𝐶 = 𝐶𝑏𝑒 + 𝐶𝑏𝑐
𝐶𝑏𝑒 ≫ 𝐶𝑏𝑐
𝐶𝑒𝑞𝐸𝐶 ≈ 𝐶𝑏𝑒
𝑓𝛽 = 𝑓𝛼(1 − 𝛼) [3.36]
𝑓𝐿 =
𝑓1
√21 𝑚⁄ − 1
[3.37]
f1 = frecuencia de corte en baja frecuencia de cada etapa.
m = número de etapas.
𝑓𝐻 = 𝑓2√2
1 𝑚⁄ − 1 [3.38]
f2 = frecuencia de corte en alta frecuencia de cada etapa.
m = número de etapas.
1
𝑓𝐻
= 1,1√
1
𝑓𝐻1
2 +
1
𝑓𝐻2
2 +⋯+
1
𝑓𝐻𝑚
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